What must be added to the polynomial 2x3 3x² 8x so that it leaves a remainder 10 when dividend by 2x 1?
Video transcript- So we have a polynomial here. What I'm curious about is what is the remainder if I were to divide this polynomial by, let's just say, x minus, I want the remainder when I divide this polynomial by x minus two? You could do this. You could figure this out with algebraic long division, but I'll give you a hint. It is much simpler and much less computation intensive and takes much less space on your paper if you use the polynomial remainder theorem. If that's unfamiliar to you, there's other videos that actually cover that. So why don't you have a go at it. All right, so now let's work through this together. The polynomial remainder theorem tells us that when I take a polynomial, p of x, and if I were to divide it by an x minus a, the remainder of that is just going to be equal to p of a. Is just going to be equal to p of a. So in this case, our p of x is this. What is our a? Well our a is going to be positive two. Remember it's x minus a. So let me do this. Our a is equal to positive two. So to figure out the remainder, we just have to evaluate p of two. So let's do that. So the remainder in this case is going to be equal to p of two, which is equal to, so let's see, it's going to be, I'll just do it all in magenta, negative three times eight minus, let's see, minus four times four plus 20 minus seven. So let's see, this is -24 minus 16 plus 20 minus seven. So that gives us, let's see, -24 minus 16, this is -40. All right, I'm just doing it step by step. This is equal to negative, actually I can do this in my head. All right, here we go. So this is -40 plus 20 is -20 minus seven is -27. That was pretty neat because if we attempted to do this without the polynomial remainder theorem, we would have had to do a bunch of algebraic long division. Now if we did the algebraic long division, we would have gotten the quotient and all of that, but we don't need the quotient, we don't need to know. So if we did all the algebraic long division, you know, we would have taken our p of x and then we would have divided the x minus a into it, and we would have gotten a quotient here, q of x, and we would have done all this business down here, all this algebraic long division. Probably wouldn't have even fit on the page. But eventually we would have gotten to a point where we got an expression that has a lower degree than this. It would have to be a constant because this is a 1st degree, so it would have be essentially a zero degree. So we would have eventually gotten to our -27. But this was much, much, much, much easier then having to go through this entire exercise. Hopefully you appreciated that. Show
Answer. So, The no. is 7. 7 must be added to the polynomial 2x^3 – 3x^2 – 8x so that it leaves a remainder 10 when divided by 2x + 1. Hence 17x – 3 is added to 6×5+5×4+11×3-3×2+x+1,so that the polynomial so obtained is exactly divisible by 3×2-2x +4. what should be added to 6x^5+4x^4-27x^3-7x^2-27x-6 so that the resulting polynomial is exactly divisible by 2x^2-3. let us divide p(x) by . thus we must add to p(x). Remainder = r(x) = Therefore, must be added to the polynomial f(x) = 6x^4 + 8x^3 + 18x^2 + 20x + 5 so that the resulting polynomial is divisible by g(x) = 3x^2 + 2x + 1. So, 2x+5 must be added to f(x), so that the result is exactly divisible by x2+x−6. Then , g(x)=0⇒x-3=0⇒x=3. By factor theorem , p (x) will be divisible by (x-3) if p (3) = 0. ⇒(15+k)=0⇒k=3. Hence , the required number to be added is 3 . Hence, 3x+5 must be subtracted from the polynomial x3+13×2+35x+25 so that the resulting polynomial is exactly divisible by x2+11x+10. Was this answer helpful? Hence, the required expression is x² – 4x + 5. Answer. 32 is the minimum number that can be substracted. Answer. 5 must be subtracted from (-1) to get (-6). 5a +b -6 should be subtracted from 2a+8b+10 to get -3a+7b+16. ML Aggarwal Solutions for Class 10 Maths Chapter 6 Factorization are provided here. Our expert faculty team has prepared solutions to help students in their exam preparation to obtain good marks in Maths. If you wish to secure an excellent score, solving ML Aggarwal Solutions is an utmost necessity. Scoring high marks requires a good amount of practice on each and every topic. These solutions will help you in gaining knowledge and a strong command over the subject. Practising the textbook questions will help you in analyzing your level of preparation and knowledge of the concept. Chapter 6 – Factorization solutions are available for download in pdf format and provide solutions to all the questions provided in ML Aggarwal Solutions for Class 10 Maths Chapter 6. Factorization is when you break a number down into smaller numbers that, multiplied together, give you that original number. When you split a number into its factors or divisors, it is called factorization. Now, let us have a look at some of the important concepts discussed in this chapter.
ML Aggarwal Solutions for Class 10 Maths Chapter 6 :-Click Here to Download PDFAccess answers to ML Aggarwal Solutions for Class 10 Maths Chapter 6 FactorizationExercise 6.1 1. Find the remainder (without division) on dividing f(x) by (x – 2) where (i) f(x) = 5x2 – 7x + 4 Solutions:- Let us assume x – 2 = 0 Then, x = 2 Given, f(x) = 5x2 – 7x + 4 Now, substitute the value of x in f(x), f(2)= (5 × 22) – (7 × 2) + 4 = (5 × 4) – 14 + 4 = 20 – 14 + 4 = 24 – 14 = 10 Therefore, the remainder is 10. (ii) f(x) = 2x3 – 7x2 + 3 Solution:- Let us assume x – 2 = 0 Then, x = 2 Given, f(x) = 2x3 – 7x2 + 3 Now, substitute the value of x in f(x), f(2)= (2 × 23) – (7 × 22) + 3 = (2 × 8) – (7 × 4) + 3 = 16 – 28 + 3 = 19 – 28 = -9 Therefore, the remainder is -9. 2. Using the remainder theorem, find the remainder on dividing f(x) by (x + 3) where (i) f(x) = 2x2 – 5x + 1 Solution:- Let us assume x + 3 = 0 Then, x = -3 Given, f(x) = 2x2 – 5x + 1 Now, substitute the value of x in f(x), f(-3)= (2 × -32) – (5 × (-3)) + 1 = (2 × 9) – (-15) + 1 = 18 + 15 + 1 = 34 Therefore, the remainder is 34. (ii) f(x) = 3x3 + 7x2 – 5x + 1 Let us assume x + 3 = 0 Then, x = -3 Given, f(x) = 3x3 + 7x2 – 5x + 1 Now, substitute the value of x in f(x), f(-3)= (3 × -33) + (7 × -32) – (5 × -3) + 1 = (3 × -27) + (7 × 9) – (-15) + 1 = – 81 + 63 + 15 + 1 = -81 + 79 = -2 Therefore, the remainder is -2. 3. Find the remainder (without division) on dividing f(x) by (2x + 1) where, (i) f(x) = 4x2 + 5x + 3 Solution:- Let us assume 2x + 1 = 0 Then, 2x = -1 X = -½ Given, f(x) = 4x2 + 5x + 3 Now, substitute the value of x in f(x), f (-½) = 4 (-½)2 + 5 (-½) + 3 = (4 × ¼) + (-5/2) + 3 = 1 – 5/2 + 3 = 4 – 5/2 = (8 – 5)/2 = 3/2 = 1½ Therefore, the remainder is 1½. (ii) f(x) = 3x3 – 7x2 + 4x + 11 Solution:- Let us assume 2x + 1 = 0 Then, 2x = -1 X = -½ Given, f(x) = 3x3 – 7x2 + 4x + 11 Now, substitute the value of x in f(x), f(-½) = (3 × (-½)3) – (7 × (-½)2 + (4 × -½) + 11 = 3 × (-1/8) – (7 × ¼) + (- 2) + 11 = -3/8 – 7/4 – 2 + 11 = – 3/8 – 7/4 + 9 = (-3 – 14 + 72)/8 = 55/8 = 4. Using remainder theorem, find the value of k if on dividing 2x3 + 3x2 – kx + 5 by x – 2 leaves a remainder 7. Solution:- Let us assume, x – 2 = 0 Then, x = 2 Given, 2x3 + 3x2 – kx + 5 Now, substitute the value of x in f(x), f(2) = (2 × 23) + (3 × 22) – (k × 2) + 5 = (2 × 8) + (3 × 4) – 2k + 5 = 16 + 12 – 2k + 5 = 33 – 2k Form the question it is given that, remainder is 7. So, 7 = 33 – 2k 2k = 33 – 7 2k = 26 K = 26/2 K = 13 Therefore, the value of k is 13. 5. Using remainder theorem, find the value of ‘a’ if the division of x3 + 5x2 – ax + 6 by (x – 1) leaves the remainder 2a. Solution:- Let us assume x -1 = 0 Then, x = 1 Given, f(x) = x3 + 5x2 – ax + 6 Now, substitute the value of x in f(x), f(1) = 13+ (5 × 12) – (a × 1) + 6 = 1 + 5 – a + 6 = 12 – a From the question it is given that, remainder is 2a So, 2a = 12 – a 2a + a = 12 3a = 12 a = 12/3 a = 4 Therefore, the value of a is 4. 6. (i) What number must be divided be subtracted from 2x2 – 5x so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1? Solution:- let us assume ‘p’ be subtracted from 2x2 – 5x So, dividing 2x2 – 5x by 2x + 1, Hence, remainder is 3 – p From the question it is given that, remainder is 2. 3 – p = 2 p = 3 – 2 p = 1 Therefore, 1 is to be subtracted. (ii) What number must be added to 2x3 – 7x2 + 2x so that the resulting polynomial leaves the remainder – 2 when divided by 2x – 3? Solution:- let us assume ‘p’ be subtracted from 2x3 – 7x2 + 2x, So, dividing it by 2x – 3, Hence, remainder is p – 6 From the question it is given that, remainder is – 2. P – 6 = – 2 P = -2 + 6 P = 4 Therefore, 4 is to be added. 7. (i) When divided by x – 3 the polynomials x3 – px2 + x + 6 and 2x3 – x2 – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’. Solution:- From the question it is given that, by dividing x3 – px2 + x + 6 and 2x3 – x2 – (p + 3)x – 6 by x – 3 = 0, then x = 3. Let us assume p(x) = x3 – px2 + x + 6 Now, substitute the value of x in p(x), p(3) = 33 – (p × 32) + 3 + 6 = 27 – 9p + 9 = 36 – 9p Then, q(x) = 2x3 – x2 – (p + 3)x – 6 Now, substitute the value of x in q(x), q(3) = (2 × 33) – (3)2 – (p + 3) × 3 – 6 = (2 × 27) – 9 – 3p – 9 – 6 = 54 – 24 – 3p = 30 – 3p Given, the remainder in each case is same, So, 36 – 9p = 30 – 3p 36 – 30 = 9p – 3p 6 = 6p p = 6/6 p = 1 Therefore, value of p is 1. (ii) Find ‘a’ if the two polynomials ax3 + 3x2 – 9 and 2x3 + 4x + a, leaves the same remainder when divided by x + 3. Solution:- Let us assume p(x) = ax3 + 3x2 – 9 and q(x) = 2x3 + 4x + a From the question it is given that, both p(x) and q(x) leaves the same remainder when divided by x + 3. Let us assume that, x + 3 = 0 Then, x = -3 Now, substitute the value of x in p(x) and in q(x), So, p(-3) = q(-3) a(-3)3 + 3(-3)2 – 9 = 2(-3)3 + 4(-3) + a -27a + 27 – 9 = – 54 – 12 + a -27a + 18 = – 66 + a -27a – a = -66 – 18 -28 a = -84 a = 84/28 Therefore, a = 3 (iii) The polynomials ax3 + 3x2 – 3 and 2x3 – 5x + a when divided by x – 4 leave the remainder r1 and r2 respectively. If 2r1 = r2, then find the value of a. Solution: Let us assume p(x) = ax3 + 3x2 – 3 and q(x) = 2x3 – 5x + a From the question it is given that, both p(x) and q(x) leaves the remainder r1 and r2 respectively when divided by x – 4. Also, given relation 2r1 = r2 Let us assume that, x – 4 = 0 Then, x = 4 Now, substitute the value of x in p(x) and in q(x), By factor theorem, r1 = p(x) and r2 = q(x) So, 2 × p(4) = q(4) 2[a(4)3 + 3(4)2 – 3] = 2(4)3 – 5(4) + a 2[64a + 48 – 3] = 128 – 20 + a 128a + 96 – 6 = 128 – 20 + a 128a + 90 = 108 + a 128a – a = 108 – 90 127a = 18 a = 18/127 Therefore, the value of a = 18/127. 8. Using remainder theorem, find the remainders obtained when x3 + (kx + 8)x + k Is divided by x + 1 and x – 2. Hence, find k if the sum of two remainders is 1. Solution: Let us assume p(x) = x3 + (kx + 8)x + k From the question it is given that, the sum of the remainders when p(x) is divided by (x + 1) and (x – 2) is 1. Let us assume that, x + 1 = 0 Then, x = -1 Also, when x – 2 = 0 Then, x = 2 Now, by remainder theorem we have p(-1) + p(2) = 1 (-1)3 + [k(-1) + 8](-1) + k + (2)3 + [k(2) + 8](2) + k = 1 -1 + k – 8 + k + 8 + 4k + 16 + k = 1 7k + 15 = 1 7k = 1 – 15 k = -14/7 k = -2 Therefore, k = -2. 9. By factor theorem, show that (x + 3) and (2x – 1) are factors of 2x2 + 5x – 3. Solution:- Let us assume, x + 3 = 0 Then, x = – 3 Given, f(x) = 2x2 + 5x – 3 Now, substitute the value of x in f(x), f(-3) = (2 × (-3)2) + (5 × -3) – 3 = (2 × 9) + (-15) – 3 = 18 – 15 – 3 = 18 – 18 = 0 Now, 2x – 1 = 0 Then, 2x = 1 x = ½ Given, f(x) = 2x2 + 5x – 3 Now, substitute the value of x in f(x), f(½) = (2 × (½)2) + (5 × ½) – 3 = (2 × (¼)) + 5/2 – 3 = ½ + 5/2 – 3 = (1 + 5)/2 – 3 = 6/2 – 3 = 3 -3 = 0 Hence, it is proved that, (x + 3) and (2x – 1) are factors of 2x2 + 5x – 3. 10. Without actual division, prove that x4 + 2x3 – 2x2 + 2x + 3 is exactly divisible by x2 + 2x – 3. Solution:- Consider x2 + 2x – 3 By factor method, x2 + 3x – x – 3 = x (x + 3) – 1(x + 3) = (x – 1) (x + 3) So, f(x) = x4 + 2x3 – 2x2 + 2x + 3 Now take, x + 3 = 0 X = -3 Then, f(-3) = (-3)4 + 2 × -(33) – (2 × (-3)2) + (2 × -3) + 3 = 81 – 54 – 18 – 6 – 3 = 0 Therefore, (x + 3) is a factor of f(x) And also, take x – 1 = 0 X = 1 Then, f(1) = 14 + 2(1)3 – 2(1)2 + 2(1) – 3 = 0 Therefore, (x – 1) is a factor of f(x) By comparing both results, p(x) is exactly divisible by x2 + 2x – 3. 11. Show that (x – 2) is a factor of 3x2 – x – 10. Hence factories 3x2 – x – 10. Solution:- Let us assume x – 2 = 0 Then, x = 2 Given, f(x) = 3x2 – x – 10 Now, substitute the value of x in f(x), f(2) = (3 × 22) – 2 – 10 = (3 × 4) – 2 – 10 = 12 – 2 – 10 = 12 – 12 = 0 Therefore, (x – 2) is a factor of f(x) Then, dividing (3x2 – x – 10) by (x – 2), we get Therefore, 3x2 – x – 10 = (x – 2) (3x + 5) 12. Using the factor theorem, show that (x – 2) is a factor of x3 + x2 – 4x – 4. Hence factorize the polynomial completely. Solution:- Let us assume, x – 2 = 0 Then, x = 2 Given, f(x) = x3 + x2 – 4x – 4 Now, substitute the value of x in f(x), f(2) = (2)3 + (2)2 – 4(2) – 4 = 8 – 4 – 8 – 4 = 0 Therefore, by factor theorem (x – 2) is a factor of x3 + x2 – 4x – 4 Then, dividing f(x) by (x – 2), we get Therefore, x3 + x2 – 4x – 4 = (x – 2) (x2 + 3x + 2) = (x – 2) (x2 + 2x + x + 2) = (x – 2) (x(x + 2) + 1(x + 2)) = (x – 2) (x + 2) (x + 1) 13. Show that 2x + 7 is a factor of 2x3 + 5x2 – 11x – 14. Hence factorize the given expression completely, using the factor theorem. Solution:- Let us assume 2x + 7 = 0 Then, 2x = -7 X = -7/2 Given, f(x) = 2x3 + 5x2 – 11x – 14 Now, substitute the value of x in f(x), f(-7/2) = 2(-7/2)3 + 5(-7/2)2 + 11(-7/2) – 14 = 2(-343/8) + 5(49/4) + (-77/2) – 14 = -343/4 + 245/4 – 77/2 – 14 = (-343 + 245 + 154 – 56)/4 = -399 + 399/4 = 0 Therefore, (2x + 7) is a factor of 2x3 + 5x2 – 11x – 14 Then, dividing f(x) by (2x + 1), we get Therefore, 2x3 + 5x2 – 11x – 14 = (2x + 7) (x2 – x – 2) = (2x + 7) (x2 – 2x + x – 2) = (2x + 7) (x(x – 2) + 1 (x – 2)) = (x + 1) (x – 2) (2x + 7) 14. Use factor theorem to factorize the following polynomials completely. (i) x3 + 2x2 – 5x – 6 Solution:- Let us assume x = -1, Given, f(x) = x3 + 2x2 – 5x – 6 Now, substitute the value of x in f(x), f(-1) = (-1)3 + 2(-1)2 – 5(-1) – 6 = -1 +2 (1) + 5 – 6 = -1 +2 + 5 – 6 = -7 + 7 = 0 Then, dividing f(x) by (x + 1), we get Therefore, x3 + 2x2 – 5x – 6 = (x + 1) (x2 + 3x – 2x – 6) = (x + 1) (x(x + 3) – 2(x + 3)) = (x + 1) (x – 2) (x + 3) (ii) x3 – 13x – 12 Solution:- Let us assume x = -1, Given, f(x) = x3 – 13x – 12 Now, substitute the value of x in f(x), f(-1) = (-1)3 – 13(-1) – 12 = -1 + 13 – 12 = – 13 + 13 = 0 Then, dividing f(x) by (x + 1), we get Therefore, x3 – 13x – 12 = (x + 1) (x2 – x – 12) = (x + 1) (x2 – 4x + 3x – 12) = (x + 1) (x(x – 4)) + 3(x – 4)) = (x + 1) (x + 3) (x – 4) 15. Use the remainder theorem to factorize the following expression. (i) 2x3 + x2 – 13x + 6 Solution:- Let us assume x = 2, Then, f(x) = 2x3 + x2 – 13x + 6 Now, substitute the value of x in f(x), f(2) = (2 × 23) + 22– 13 × 2 + 6 = (2 × 8) + 4 – 26 + 6 = 16 + 4 – 26 + 6 = 26 – 26 = 0 Then, dividing f(x) by (x – 2), we get Therefore, 2x3 + x2 – 13x + 6 = (x – 2) (2x2 + 5x – 3) = (x – 2)(2x2 + 6x – x – 3) = (x – 2) (2x(x + 3) – 1 (x + 3)) = (x – 2) (x + 3) (2x – 1) (ii) 3x2 + 2x2 – 19x + 6 Solution:- Given, f(x) = 3x3 + 2x2 – 19x + 6 Let us assume x = 1 Then, f(1) = 3(1)3 + 2(1)2 – (19 × 1) + 6 = 3 + 2 – 19 + 6 = 11 – 19 = – 8 So, – 8 ≠ 0 Let us assume x = -1 Then, f(-1) = 3(-1)3 + 2(-1)2 – (19 × (-1)) + 6 = – 3 + 2 + 19 + 6 = – 3 + 27 = 24 So, 24 ≠ 0 Now, assume x = 2 Then, f(2) = 3(2)3 + 2(2)2 – (19 × (2)) + 6 = 24 + 8 – 38 + 6 = 38 – 38 = 0 So, 0 = 0 Therefore, (x – 2) is a factor of f(x). f(x) = 3x3 + 2x2 – 19x + 6 = 3x3 – 6x2 + 8x2 – 16x – 3x + 6 = 3x2 (x – 2) + 8x (x – 2) – 3(x – 2) = (x – 2) (3x2 + 8x – 3) = (x – 2) (3x2 + 9x – x – 3) = (x – 2) (3x(x + 3) – 1(x + 3) = (x – 2) (x + 3) (3x – 1) (iii) 2x3 + 3x2 – 9x – 10 Solution:- Given, f(x) = 2x3 + 3x2 – 9x – 10 Let us assume, x = -1 = 2(-1)3 + 3(-1)2 – 9 (-1) – 10 = -2 + 3 + 9 – 10 = 12 – 12 = 0 Therefore, (x + 1) is the factor of 2x3 + 3x2 – 9x – 10 Then, dividing f(x) by (x + 1), we get Therefore, 2x3 + 3x2 – 9x – 10 = 2x2 + 5x – 4x – 10 = x(2x + 5) – 2 (2x + 5) – (2x + 5) (x – 2) Hence the factors are (x + 1) (x – 2) (2x + 5) (iv) x3 + 10x2 – 37x + 26 Solution:- Given, f(x) = x3 + 10x2 – 37x + 26 Let us assume, x = 1 Then, f(1) = 13 + 10(1)2 – 37 (1) + 26 = 1 + 10 – 37 + 26 = 37 – 37 = 0 Therefore, x – 1is a factor of x3 + 10x2 – 37x + 26 Then, dividing f(x) by (x – 1), we get Therefore, x3 + 10x2 – 37x + 26 = (x – 1) (x2 + 11x – 26) = (x – 1) (x2 + 13x – 2x – 26) = (x – 1) (x (x + 13) – 2(x + 13)) = (x – 1) ((x – 2) (x + 13)) 16. If (2x + 1) is a factor of 6x3 + 5x2 + ax – 2 find the value of a. Solution:- Let us assume 2x + 1 = 0 Then, 2x = – 1 X = -½ Given, f(x) = 6x3 + 5x2 + ax – 2 Now, substitute the value of x in f(x), f (-½) = 6 (-½)3 + 5 (-½)2 + a (-½) – 2 = 6 (-1/8) + 5 (¼) – ½a – 2 = -3/4 + 5/4 – a/2 – 2 = (-3 + 4 – 2a – 8)/4 = (-6 – 2a)/4 From the question, (2x + 1) is a factor of 6x3 + 5x2 + ax – 2 Then, remainder is 0. So, (-6 – 2a)/4 = 0 -6 – 2a = 4 × 0 – 6 – 2a = 0 -2a = 6 a = -6/2 a = – 3 Therefore, the value of a is – 3. 17. If (3x – 2) is a factor of 3x3 – kx2 + 21x – 10, find the value of k. Solution:- Let us assume 3x – 2 = 0 Then, 3x = 2 X = 2/3 Given, f(x) = 3x3 – kx2 + 21x – 10 Now, substitute the value of x in f(x), f (2/3) = 3 (2/3)3 – k (2/3)2 + 21 (2/3) – 10 = 3 (8/27) – k (4/9) + 14 – 10 = 8/9 – 4k/9 + 14 – 10 = 8/9 – 4k/9 + 4 = (8 – 4k + 36)/9 = (44 – 4k)/9 From the question, (3x – 2) is a factor of 3x3 – kx2 + 21x – 10 Then, remainder is 0 So, (44 – 4k)/9 = 0 44 – 4k = 0 × 9 44 = 4k K = 44/4 K = 11 18. If (x – 2) is a factor of 2x3 – x2 + px – 2, then (i) find the value of p. (ii) with this value of p, factorize the above expression completely. Solution:- Let us assume x -2 = 0 Then, x = 2 Given, f(x) = 2x3 – x2 + px – 2 Now, substitute the value of x in f(x), f(2) = (2 × 23) – 22 + (p × 2) – 2 = (2 × 8) – 4 + 2p – 2 = 16 – 4 + 2p – 2 = 16 – 6 + 2p = 10 + 2p From the question, (x – 2) is a factor of 2x3 – x2 + px – 2 Then, remainder is 0. 10 + 2p = 0 2p = – 10 P = -10/2 P = -5 So, (x – 2) is a factor of 2x3 – x2 + 5x – 2 Therefore, 2x3 – x2 + 5x – 2 = (x – 2) (2x2 + 3x + 1) = (x – 2) (2x2 + 2x + x + 1) = (x – 2) (2x(x + 1) + 1(x + 1)) = (x + 1) (x – 2) (2x + 1) 19. What number should be subtracted from 2x3 – 5x2 + 5x so that the resulting polynomial has 2x – 3 as a factor? Solution:- Let us assume the number to be subtracted from 2x3 – 5x2 + 5x be p. Then, f(x) = 2x3 – 5x2 + 5x – p Given, 2x – 3 = 0 x = 3/2 f(3/2) = 0 So, f(3/2) = 2(3/2)3 – 5(3/2)2 + 5(3/2) – p = 0 2(27/8) – 5(9/4) + 15/2 – p = 0 27/4 – 45/4 + 15/2 – p = 0 [multiply by 4 for all numerator] 27 – 45 + 30 – 4p = 0 57 – 45 – 4p = 0 12 – 4p = 0 P = 12/4 P = 3 Therefore, 3 is the number should be subtracted from 2x3 – 5x2 + 5x. 20. (i) Find the value of the constants a and b, if (x – 2) and (x + 3) are both factors of the expression x3 + ax2 + bx – 12. Solution:- Let us assume x – 2 = 0 Then, x = 2 Given, f(x) = x3 + ax2 + bx – 12 Now, substitute the value of x in f(x), f(2) = 23 + a(2)2 + b(2) – 12 = 8 + 4a + 2b – 12 = 4a + 2b – 4 From the question, (x – 2) is a factor of x3 + ax2 + bx – 12. So, 4a + 2b – 4 = 0 4a + 2b = 4 By dividing both the side by 2 we get, 2a + b = 2 … [equation (i)] Now, assume x + 3 = 0 Then, x = -3 Given, f(x) = x3 + ax2 + bx – 12 Now, substitute the value of x in f(x), f(-3) = (-3)3 + a(-3)2 + b(-3) – 12 = -27 + 9a – 3b – 12 = 9a – 3b – 39 From the question, (x – 3) is a factor of x3 + ax2 + bx – 12. So, 9a – 3b – 39 = 0 9a – 3b = 39 By dividing both the side by 3 we get, 3a – b = 13 … [equation (ii)] Now, adding both equation (i) and equation (ii) we get, (2a + b) + (3a – b) = 2 + 13 2a + 3a + b – b = 15 5a = 15 a = 15/5 a = 3 Consider the equation (i) to find out ‘b’. 2a + b = 2 2(3) + b = 2 6 + b = 2 b = 2 – 6 b = -4 (ii) If (x + 2) and (x + 3) are factors of x3 + ax + b, Find the values of a and b. Solution:- Let us assume x + 2 = 0 Then, x = -2 Given, f(x) = x3 + ax + b Now, substitute the value of x in f(x), f(-2) = (-2)3 + a(-2) + b = -8 – 2a + b From the question, (x + 2) is a factor of x3 + ax + b. Therefore, remainder is 0. f(x) = 0 – 8 – 2a + b = 0 2a – b = – 8 … [equation (i)] Let us assume x + 3 = 0 Then, x = -3 Given, f(x) = x3 + ax + b Now, substitute the value of x in f(x), f(-2) = (-3)3 + a(-3) + b = -27 – 3a + b From the question, (x + 3) is a factor of x3 + ax + b. Therefore, remainder is 0. f(x) = 0 – 27 – 3a + b = 0 3a – b = – 27 … [equation (i)] Now, subtracting both equation (i) and equation (ii) we get, (2a – b) – (3a – b) = -8 – (-27) 2a – 3a – b + b = – 8 + 27 -a = 19 a = -19 Consider the equation (i) to find out ‘b’. 2a – b = – 8 2(-19) – b = -8 -38 – b = – 8 b = -38 +8 b = -30 21. If (x + 2) and (x – 3) are factors of x3 + ax + b, find the values of a and b. With these values of a and b, factorize the given expression. Solution:- Let us assume x + 2 = 0 Then, x = -2 Given, f(x) = x3 + ax + b Now, substitute the value of x in f(x), f(-2) = (-2)3 + a(-2) + b = -8 – 2a + b From the question, (x + 2) is a factor of x3 + ax + b. Therefore, remainder is 0. f(x) = 0 – 8 – 2a + b = 0 2a – b = – 8 … [equation (i)] Now, assume x – 3 = 0 Then, x = 3 Given, f(x) = x3 + ax + b Now, substitute the value of x in f(x), f(3) = (3)3 + a(3) + b = 27 + 3a + b From the question, (x – 3) is a factor of x3 + ax + b. Therefore, remainder is 0. f(x) = 0 27 + 3a + b = 0 3a + b = – 27 … [equation (ii)] Now, adding both equation (i) and equation (ii) we get, (2a – b) + (3a + b) = – 8 – 27 2a – b + 3a + b = -35 5a = -35 a = -35/5 a = -7 Consider the equation (i) to find out ‘b’. 2a – b = – 8 2(-7) – b = -8 -14 – b = -8 b = – 14 + 8 b = -6 Therefore, value of a = -7 and b = -6. Then, f(x) = x3 – 7x – 6 (x + 2) (x – 3) = x(x – 3) + 2(x – 3) = x2 – 3x + 2x – 6 = x2 – x – 6 Dividing f(x) by x2 – x – 6 we get, Therefore, x3 – 7x – 6 = (x + 1) (x + 2) (x – 3) 22. (x – 2) is a factor of the expression x3 + ax2 + bx + 6. When this expression is divided by (x – 3), it leaves the remainder 3. Find the values of a and b. Solution:- From the question it is given that, (x – 2) is a factor of the expression x3 + ax2 + bx + 6 Then, f(x) = x3 + ax2 + bx + 6 … [equation (i)] Let assume x – 2 = 0 Then, x = 2 Now, substitute the value of x in f(x), f(2) = 23+ a(2)2 + 2b + 6 = 8 + 4a + 2b + 6 = 14 + 4a + 2b By dividing the numbers by 2 we get, = 7 + 2a + b From the question, (x – 2) is a factor of the expression x3 + ax2 + bx + 6. So, remainder is 0. f(x) = 0 7 + 2a + b = 0 2a + b = -7 … [equation (ii)] Now, expression is divided by (x – 3), it leaves the remainder 3. So, remainder = 33 + 9a + 3b = 3 9a + 3b = 3 – 33 9a + 3b = -30 By dividing the numbers by 3 we get, = 3a + b = – 10 … [equation (iii)] Now, subtracting equation (iii) from equation (ii) we get, (3a + b) – (2a + b) = – 10 – (-7) 3a – 2a + b – b = – 10 + 7 a = -3 Consider the equation (ii) to find out ‘b’. 2a + b = – 7 2(-3) + b = – 7 -6 + b = – 7 b = – 7 + 6 b = – 1 23. If (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 and when the expression is divided by (x – 3), it leaves a remainder 52, find the values of a and b. Solution:- From the question it is given that, (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14 Then, f(x) = 2x3 + ax2 + bx – 14 … [equation (i)] Let assume x – 2 = 0 Then, x = 2 Now, substitute the value of x in f(x), f(2) = 2(2)3+ a(2)2 + 2b – 14 = 16 + 4a + 2b – 14 = 2 + 4a + 2b By dividing the numbers by 2 we get, = 1 + 2a + b From the question, (x – 2) is a factor of the expression 2x3 + ax2 + bx – 14. So, remainder is 0. f(x) = 0 1 + 2a + b = 0 2a + b = -1 … [equation (ii)] Now, expression is divided by (x – 3), it leaves the remainder 52. So, remainder = 9a + 3b + 40 = 52 9a + 3b = 52 – 40 9a + 3b = 12 By dividing the numbers by 3 we get, = 3a + b = 4 … [equation (iii)] Now, subtracting equation (iii) from equation (ii) we get, (3a + b) – (2a + b) = 4 – (-1) 3a – 2a + b – b = 4 + 1 a = 5 a = 5 Consider the equation (ii) to find out ‘b’. 2a + b = – 1 2(5) + b = – 1 10 + b = – 1 b = – 1 – 10 b = – 11 24. If ax3 + 3x2 + bx – 3 has a factor (2x + 3) and leaves remainder – 3 when divided by (x + 2), find the values of a and b. With these values of a and b, factorize the given expression. Solution:- Let us assume, 2x + 3 = 0 Then, 2x = -3 x = -3/2 Given, f(x) = ax3 + 3x2 + bx – 3 Now, substitute the value of x in f(x), f(-3/2) = a(-3/2)3 + 3(-3/2)2 + b(-3/2) – 3 = a(-27/8) + 3(9/4) – 3b/2 – 3 = -27a/8 + 27/4 – 3b/2 – 3 From the question it is given that,ax3 + 3x2 + bx – 3 has a factor (2x + 3). So, remainder is 0. -27a/8 + 27/4 – 3b/2 – 3 = 0 -27a + 54 – 12b – 24 = 0 -27a – 12b = -30 By dividing the numbers by – 3 we get, 9a + 4b = 10 [equation (i)] Now, let us assume x + 2 = 0 Then, x = -2 Given, f(x) = ax3 + 3x2 + bx – 3 Now, substitute the value of x in f(x), f(2) = a(-2)3 + 3(-2)2 + b(-2) – 3 = -8a + 12 – 2b – 3 = -8a – 2b + 9 Leaves the remainder -3 So, -8a – 2b + 9 = -3 -8a – 2b = -3 – 9 -8a – 2b = -12 By dividing both sides by -2 we get, 4a + b = 6 [equation (ii)] By multiplying equation (ii) by 4, 16a + 4b = 24 Now, subtracting equation (ii) from equation (i) we get, (16a + 4b) – (9a + 4b) = 24 – 10 16a – 9a + 4b – 4b = 14 7a = 14 a = 14/7 a = 2 Consider the equation (i) to find out ‘b’. 9a + 4b = 10 9(2) + 4b = 10 18 + 4b = 10 4b = 10 – 18 4b = -8 b = -8/4 b = -2 Therefore, f(x) = ax3 + 3x2 + bx – 3 = 2x3 + 3x2 – 2x – 3 Given, 2x + 3 is a factor of f(x) So, divide f(x) by 2x + 3 Therefore, 2x3 + 3x2 – 2x – 3 = (2x + 3) (x2 – 1) = (2x + 3) (x + 1) (x – 1) 25. Given f(x) = ax2 + bx + 2 and g(x) = bx2 + ax + 1. If x – 2 is a factor of f(x) but leaves the remainder – 15 when it divides g(x), find the values of a and b. With these values of a and b, factorize the expression. f(x) + g(x) + 4x2 + 7x. Solution:- From the question it is given that, f(x) = ax2 + bx + 2 and g(x) = bx2 + ax + 1 and x – 2 is a factor of f(x), So, x = 2 Now, substitute the value of x in f(x), f(2) = 0 a(2)2 + b(2) + 2 = 0 4a + 2b + 2 = 0 By dividing both sides by 2 we get, 2a + b + 1 = 0 … [equation (i)] Given, g(x) divide by (x – 2), leaves remainder – 15 g(x) = bx2 + ax + 1 So, g(2) = -15 b(2)2 + 2a + 1 = -15 4b + 2a + 1 + 15 = 0 4b + 2a + 16 = 0 By dividing both sides by 2 we get, 2b + a + 8 = 0 … [equation (ii)] Now, subtracting equation (ii) from equation (i) multiplied by 2, (4a + 2b + 2) – (a + 2b + 8) = 0 – 0 4a – a + 2b – 2b + 2 – 8 = 0 3a – 6 = 0 3a = 6 a = 6/3 a = 2 Consider the equation (i) to find out ‘b’. 2a + b + 1 = 0 2(2) + b = – 1 4 + b = – 1 b = – 1 – 4 b = – 5 Now, f(x) = ax2 + bx + 2 = 2x2 – 5x + 2 g(x) = bx2 + ax + 1 = -5x2 + 2x + 1 then, f(x) + g(x) + 4x2 + 7x = 2x2 – 5x + 2 – 5x2 + 2x + 1 + 4x2 + 7x = x2 + 4x + 3 = x2 + 3x + x + 3 = x(x + 3) + 1(x + 3) = (x + 1) (x + 3) Chapter Test 1. Find the remainder when 2x3 – 3x2 + 4x + 7 is divided by (i) x – 2 (ii) x + 3 (iii) 2x + 1 Solution:- From the question it is given that, f(x) = 2x3 – 3x2 + 4x + 7 (i) Consider x -2 let us assume x – 2 = 0 Then, x = 2 Now, substitute the value of x in f(x), f(2) = 2(2)3 – 3(2)2 + 4(2) + 7 = 16 – 12 + 8 + 7 = 31 -12 = 19 Therefore, the remainder is 19 (ii) consider x + 3 let us assume x + 3 = 0 Then, x = -3 Now, substitute the value of x in f(x), f(2) = 2(-3)3 – 3(-3)2 + 4(-3) + 7 = 2(-27) – 3(9) – 12 + 7 = – 54 – 27 – 12 + 7 = – 93 + 7 = – 86 Therefore, remainder is -86. (iii) consider 2x + 1 Let us assume, 2x + 1 = 0 Then, 2x = -1 X = -½ Now, substitute the value of x in f(x), f (-½) = 2 (-½)3 – 3(-½)2 + 4 (-½) + 7 = 2(-1/8) – 3 (¼) + 4 (-½) + 7 = -¼ – ¾ – 2 + 7 = -1 – 2 + 7 = 4 Therefore, remainder is 4. 2. When 2x3 – 9x2 + 10x – p is divided by (x + 1), the remainder is – 24. Find the value of p. Solution:- Let us assume x + 1 = 0 Then, x = -1 Given, f(x) = 2x3 – 9x2 + 10x – p Now, substitute the value of x in f(x), f(-1) = 2(-1)3 – 9(-1)2 + 10(-1) – p = -2 – 9 – 10 + p = -21 + p From the question it is given that, the remainder is – 24, So, -21 + p = -24 p = – 24 + 21 p = -3 So, f(x) = 2x3 – 9x2 + 10x – (-3) = 2x3 – 9x2 + 10x + 3 Therefore, the value of p is 3. 3. If (2x – 3) is a factor of 6x2 + x + a, find the value of a. With this value of a, factorise the given expression. Solution:- Let us assume 2x – 3 = 0 Then, 2x = 3 X = 3/2 Given, f(x) = 6x2 + x + a Now, substitute the value of x in f(x), f(3/2) = 6(3/2)2 + (3/2) + a = 6(9/4) + (3/2) + a = 3(9/2) + (3/2) + a = 27/2 + 3/2 + a = 30/2 + a = 15 + a From the question, (2x – 3) is a factor of 6x2 + x + a. So, remainder is 0. Then, 15 + a = 0 a = -15 Therefore, f(x) = 6x2 + x – 15 Dividing f(x) by 2x – 3 we get, Therefore, 6x2 + x – 15 = (2x – 3) (3x + 5) 4. When 3x2 – 5x + p is divided by (x – 2), the remainder is 3. Find the value of p. Also factorize the polynomial 3x2 – 5x + p – 3. Solution:- Let us assume x – 2 = 0 Then, x = 2 Given, f(x) = 3x2 – 5x + p Now, substitute the value of x in f(x), So, f(2) = 3(2)2 – 5(2) + p = 3(4) – 10 + p = 12 – 10 + p = 2 + p From the question it is given that, remainder is 3. So, 2 + p = 3 p = 3 – 2 p = 1 Therefore, f(x) = 3x2 – 5x + 1 Consider the polynomial, 3x2 – 5x + p – 3 Now, substitute the value of p in polynomial, = 3x2 – 5x + 1 – 3 = 3x2 – 5x – 2 Now, by factorizing the polynomial 3x2 – 5x – 2, Dividing 3x2 – 5x – 2 by x – 2 we get, Therefore, 3x2 – 5x – 2 = (x – 2) (3x + 1) 5. Prove that (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. Hence factorize the given polynomial completely. Solution:- Let us assume (5x + 4) = 0 Then, 5x = -4 x = -4/5 Given, f(x) = 5x3 + 4x2 – 5x – 4 Now, substitute the value of x in f(x), So, f(-4/5) = 5(-4/5)3 + 4(-4/5)2 – 5(-4/5) – 4 = 5(-64/125) + 4 (16/25) + 4 – 4 = -64/25 + 64/25 = (-64 + 64)/25 = 0/25 = 0 Hence, (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. So, dividing 5x3 + 4x2 – 5x – 4 by 5x + 4 we get, Therefore, 5x3 + 4x2 – 5x – 4 = (5x + 4) (x2 – 1) = (5x + 4) (x2 – 12) = (5x + 4) (x + 1) (x – 1) 6. Use factor theorem to factorize the following polynomials completely: (i) 4x3 + 4x2 – 9x – 9 Solution:- Let us assume x = -1, Given, f(x) = 4x3 + 4x2 – 9x – 9 Now, substitute the value of x in f(x), f(-1) = 4(-1)3 + 4(-1)2 – 9(-1) – 9 = -4 + 4 + 9 – 9 = 0 Therefore, x + 1 is the factor of 4x3 + 4x2 – 9x – 9. Now, dividing 4x3 + 4x2 – 9x – 9 by x + 1 we get, Therefore, 4x3 + 4x2 – 9x – 9 = (x + 1) (4x2 – 9) = (x + 1) ((2x)2 – (3)2) = (x + 1) (2x + 3) (2x – 3) (ii) x3 – 19x – 30 Solution:- Let us assume x = -2, Given, f(x) = x3 – 19x – 30 Now, substitute the value of x in f(x), f(-1) = (-2)3 – 19(-2) – 30 = -8 + 38 – 30 = -38 + 38 = 0 Therefore, x + 2 is the factor of x3 – 19x – 30. Now, dividing x3 – 19x – 30 by x + 2 we get, Therefore, x3 – 19x – 30 = (x + 2)(x2 – 2x – 15) = (x + 2) (x2 – 5x + 3x – 15) = (x + 2) (x – 5) (x + 3) 7. If x3 – 2x2 + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorize the given polynomial completely. Solution:- From the question it is given that, (x + 2) is a factor of the expression x3 – 2x2 + px + q Then, f(x) = x3 – 2x2 + px + q Let assume x + 2 = 0 Then, x = -2 Now, substitute the value of x in f(x), f(-2) = (-2)3 – 2(-2)2 + p(-2) + q = -8 – 8 – 2p + q = -16 – 2p + q 2p – q = – 16 … [equation (i)] Now, consider (x + 1) Then, f(x) = x3 – 2x2 + px + q Let assume x + 1 = 0 Then, x = -1 Now, substitute the value of x in f(x), f(-1) = (-1)3 – 2(-1)2 + p(-1) + q = -1 – 2 –p + q = – 3 – p + q Given, remainder is 9 So, -3 – p + q = 9 – p + q = 9 + 3 -p + q = 12 … [equation (ii)] Now, adding equation (i) and equation (ii) we get, (2p – q) + (-p + q) = – 16 + 12 2p – q – p + q = -4 P = -4 Consider the equation (ii) to find out ‘b’. – p + q = 12 -(-4) + q = 12 4 + q = 12 q = 12 – 4 q = 8 Therefore, by substituting the value of p and q f(x) = x3 – 2x2 – 4x + 8 Dividing f(x) be (x + 2) we get, x3 – 2x2 – 4x + 8 = (x + 2) (x2 – 4x + 4) = (x + 2) (x2 – 2 × x (-2) + 22) = (x + 2) (x – 2)2 8. If (x + 3) and (x – 4) are factors of x3 + ax2 – bx + 24, find the values of a and b: With these values of a and b, factorize the given expression. Solution:- Let us assume x + 3 = 0 Then, x = -3 Given, f(x) = x3 + ax2 – bx + 24 Now, substitute the value of x in f(x), f(-3) = (-3)3 + a(-3)2 – b(-3) + 24 = -27 + 9a + 3b + 24 = 9a + 3b – 3 Dividing all terms by 3 we get, = 3a + b – 1 From the question, (x + 3) is a factor of x3 + ax2 – bx + 24. Therefore, remainder is 0. f(x) = 0 3a + b – 1 = 0 3a + b = 1 … [equation (i)] Now, assume x – 4 = 0 Then, x = 4 Given, f(x) = x3 + ax2 – bx + 24 Now, substitute the value of x in f(x), f(4) = 43 + a(4)2 – b(4) + 24 = 64 + 16a – 4b + 24 = 88 + 16a – 4b Dividing all terms by 4 we get, = 22 + 4a – b From the question, (x – 4) is a factor of x3 + ax2 – bx + 24. Therefore, remainder is 0. f(x) = 0 22 + 4a – b = 0 4a – b = – 22 … [equation (ii)] Now, adding both equation (i) and equation (ii) we get, (3a + b) + (4a – b) = 1 – 22 3a + b + 4a – b = – 21 7a = – 21 a = -21/7 a = -3 Consider the equation (i) to find out ‘b’. 3a + b = 1 3(-3) + b = 1 -9 + b = 1 b = 1 + 9 b = 10 Therefore, value of a = -3 and b = 10. Then, by substituting the value of a and b f(x) = x3 – 3x2 – 10x + 24 (x + 3) (x – 4) = x(x – 4) + 3(x – 4) = x2 – 4x + 3x – 12 = x2 – x – 12 Dividing f(x) by x2 – x – 12 we get, Therefore, x3 – 3x2 – 10x + 24 = (x2 – x – 12) (x – 2) = (x + 3) (x – 4) (x – 2) 9. If (2x + 1) is a factor of both the expressions 2x2 – 5x + p and 2x2 + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials. Solution:- Let us assume 2x + 1 = 0 Then, 2x = -1 x = -½ Given, p(x) = 2x2 – 5x + p Now, substitute the value of x in p(x), p (-½) = 2 (-½)2 – 5(-½) + p = 2(1/4) + 5/2 + p = ½ + 5/2 + p = 6/2 + p = 3 + p From the question it is given that, (2x + 1) is a factor of both the expressions 2x2 – 5x + p So, remainder is 0. Then, 3 + p = 0 p = – 3 Now consider q(x) = 2x2 + 5x + q Substitute the value of x in q(x) q (-½) = 2 (-½)2 + 5(-½) + q = 2(1/4) – 5/2 + q = ½ – 5/2 + q = (1 – 5)/2 + q = -4/2 + q = q – 2 From the question it is given that, (2x + 1) is a factor of both the expressions 2x2 + 5x + q So, remainder is 0. q – 2 = 0 q = 2 Therefore, p = – 3 and q = 2 P(x) = 2x2 – 5x – 3 q(x) = 2x2 + 5x + 2 Then, divide p(x) by 2x + 1 Therefore, 2x2 – 5x – 3 = (2x + 1) (x – 3) Now, divide q(x) by 2x + 1 Therefore, 2x2 + 5x + 2 = (2x + 1) (x + 2) 10. If a polynomial f(x)= x4-2x3 + 3x2– ax + b leaves reminder 5 and 19 when divided by (x – 1) and (x + 1) respectively, Find the values of a and b. Hence determined the reminder when f(x) is divided by (x-2). Solution:- From the question it is given that, f(x) = x4 – 2x3 +3x2 – ax + b Factor (x – 1) leaves remainder 5, Factor (x + 1) leaves remainder 19, Where x = 1 and x = – 1 f(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + b = 19 1 – 2(-1) + 3(1) – a(-1) + b = 19 1 + 2 + 3 + a + b = 19 6 + a + b = 19 a + b = 19 – 6 a + b = 13 … [equation (i)] f(1) = (1)4 – 2(1)3 + 3(1)2 – a(1) + b = 5 1 – 2(1) + 3(1) – a(1) + b = 5 1 – 2 + 3 – a + b = 5 2 – a + b = 5 – a + b = 5 – 2 – a + b = 3 … [equation (ii)] Now, subtracting equation (ii) from equation (i) we get, (a + b) – (- a + b) = 13 – 3 a + b + a – b = 10 2a = 10 a = 10/2 a = 5 To find out the value of b, substitute the value of a in equation (i) we get, a + b = 13 5 + b = 13 b = 13 – 5 b = 8 Therefore, value of a = 5 and b = 8 11. When a polynomial f(x) is divided by (x – 1), the remainder is 5 and when it is, divided by (x – 2), the remainder is 7. Find the remainder when it is divided by (x – 1) (x – 2). Solution:- From the question it is given that, Polynomial f(x) is divided by (x – 1), Remainder = 5 Let us assume x – 1 = 0 x = 1 f(1) = 5 and the divided be (x – 2), remainder = 7 let us assume x – 2 = 0 x = 2 Therefore, f(2) = 7 So, f(x) = (x – 1) (x – 2) q(x) + ax + b Where, q(x) is the quotient and ax + b is remainder, Now put x = 1, we get, f(1) = (1 – 1)(1 – 2)q(1) + (a × 1) + b a + b = 5 … [equation (i)] x = 2, f(2) = (2 – 1)(2 – 2)q(2) + (a × 2) + b 2a + b = 7 … [equation (ii)] Now subtracting equation (i) from equation (ii) we get, (2a + b) – (a + b) = 7 – 5 2a + b – a – b = 2 a = 2 To find out the value of b, substitute the value of a in equation (i) we get, a + b = 5 2 + b = 5 b = 5 – 2 b = 3 Therefore, the remainder = ax + b = 2x + 3 Thus, if we subtract remainder from the dividend, then it will be exactly divisible by the divisor. Therefore the required polynomial to be subtracted from 2 x 4 - 11 x 3 + 29 x 2 - 40 x + 29 so that it is exactly divisible by x 2 - 3 x + 4 is - 2 x + 5 . 1 Answer. Required number to be subtracted = 5.
Answer: 7 must be added. So, The no. is 7. 7 must be added to the polynomial 2x^3 - 3x^2 - 8x so that it leaves a remainder 10 when divided by 2x + 1. Thus, the number that must be added would be 7. What must be added to the polynomial 2x3 3x2 8x so that it leaves a remainder 10 when divided by 2x 1?Answer: 7 must be added.
What must be added to the polynomial 2x3 so that it leaves a remainder 10 when divided by 2x 1?7 must be added to the polynomial 2x^3 - 3x^2 - 8x so that it leaves a remainder 10 when divided by 2x + 1. Thus, the number that must be added would be 7.
What should be added in 2x +1 to get the result as 1?-2x should be added to 2x+1 to get the result 1.. Given, An expression 2x+1.. To Find, What should be added in 2x+1 to get the result as 1.. Solution, Let us assume that on the addition of a in 2x+1 the result is obtained as 1. The given expression is. 2x+1. So, 2x+1+a = 1. Now, taking 1 to R.H.S. 2x+a = 0.. |
