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The Polynomial F(X) = Ax4 + X3 + Bx2 - 4x + C Has (X + 1), (X-2) and (2x - 1) as Its Factors - Mathematics

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Question

The polynomial f(x) = ax4 + x3 + bx2 - 4x + c has (x + 1), (x-2) and (2x - 1) as its factors. Find the values of a,b,c, and remaining factor. 

Sum

Solution

When x + 1 is a factor, we can substitute x= -1 to evaluate values .... (i) 

When x - 2 is a factor, we can substitute x = 2 to evaluate values .... (ii) 

When 2x - 1 is a factor, we can substitute  x = `1/2` to evaluate values ....(iii)

Substituting  (1), we get 

f(-1) = a × (-1)4 + (-1)3 + b(-1)2 - 4(-1) + c = 0

⇒ a + b + c = -3 

⇒ a = - b - c - 3 ....(iv)

Substituting (11 ), we get 

⇒ f(2) =  a × (2)4 + (2)3 + b(2)2 - 4(2) + c = O 

⇒ 16 a + 4 b  + c = 0  .....(v)

Substituting (iii), we get 

⇒ `"f"(1/2) = "a" xx (1/2)^4 + (1/2)^3 + "b"(1/2)^2 - 4(1/2) + c = 0`

`=> "a"/16 + "b"/4 + "c" = 2 -1/8`

⇒ a + 4b + 16c = 30  ..(vi)

Putting (iv) in (v) and (vi), we get: 

16a + 4b + c = O; ⇒ 16x(- b - c - 3) + 4b+ c = O 

⇒ -12 b - 15 c - 48 = 0 = 4b + 5c = 16 

⇒ b = - 4 - `"5c"/4`  .......(vii)

a + 4b + 16c = 30 ⇒ (- b- c - 3) + 4b + 16c = 30 

⇒ 3b + 15 c = 33 .... (viii)

Putting (vii) 1n (viii) ,  we get,  

⇒ 3 x `(-4 - (5"c")/4) + 15"c" = 33`,

⇒ Solving this , we get

⇒  c = 4

Putting this value of c in (viii), we get: 

b = -9

Putting this value of c in (iv), we get: 

a = 2

Puting values of a,b,c in polynomial , we get :

f (x) = 2x4 + x3 - 9x2 - 4x + 4

In order to find the remaining factor, lets start with (x-2) as one of the factor. Then, 

Multiplying (x - 2) by 2x3, we get 2x4 - 4x3, hence we are left with 5x3 -9x2 - 4x + 4 (and 1st part of factor as 2x3). 

Multiplying (x-2) by 5x2, we get 5x3 - 10x2, hence we are left with x2 - 4x + 4 (and 2"d part of factor as 5x2). 

Multiplying (x-2) by x, we get x2 - 2x, hence we are left with -2x + 4 (and 3'd part of factor as x). 

Multiplying (x - 2) by -2, we get -2x + 4, hence we are left with 0 (and 4th part of factor as -2). 

Hence complete factor is (x - 2)(2x3 + 5x2 + x - 2). 

Again factoring (2x3 + 5x2 + x - 2) by similar method we get factors as: 

(x +1 )(2x - 1)(x + 2) 

Hence, f(x) = 2x4 + x3 - 9x2 - 4x +4 = (x - 2)(x +1)(2x - 1)(x + 2).

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Chapter 10: Remainder And Factor Theorems - Exercise 10.1

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Frank Mathematics - Part 2 [English] Class 10 ICSE
Chapter 10 Remainder And Factor Theorems
Exercise 10.1 | Q 22

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