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X4 − 2x3 − 7x2 + 8x + 12 - Mathematics

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Question

x4 − 2x3 − 7x2 + 8x + 12

Answer in Brief

Solution

Let  `f(x) = x^4 - 2x^3 - 7x^2 + 8x + 12` be the given polynomial.

Now, putting  x = -1we get

`f (-1) = (-1)^4 -2 (-1)^3 - 7 ( -1)^2 + 8 (-1)+12`

             ` = 1 + 2 - 7 - 8 _ 12 = -15 + 15`

             ` = 0`

Therefore, (x + 1)is a factor of polynomial f(x).

Now,

\[f(x) = x^4 - 3 x^3 + x^3 - 3 x^2 - 4 x^2 + 12x - 4x + 12\]

`f(x) = x^3 (x+1) -3x^2 (x +1) - 4x(x+1) + 12 (x +1)`

` = (x +1){x^3 -3x^2 - 4x + 12}`

`=(x+1)g(x)      ......... (1)`

Where `g(x)=x^3 -3x^2 - 4x + 12`

Putting  x = 2,we get

`g(2) = (2)^3 -3(2)^2 - 4 (2) + 12`

`8 -12 -8 +12 = 20 -20`

                             ` = 0`

Therefore, (x -2)is the factor of g(x).

Now,

\[g(x) = x^3 - 2 x^2 - x^2 - 6x + 2x + 12\]

`g(x) = x^2 (x -2) -x (x -2) -6(x - 2)`

` =  (x -2){x^2 -x -6}`

` = (x-2){x^2 - 3x +2x -6}`

` = (x-2)(x+2) (x-3)    ......... (2)`

From equation (i) and (ii), we get

 `f(x) = (x+1)(x -2)(x+2)(x-3)`

Hence (x +1),(x -2),(x +2) and (x-3) are the factors of polynomial f(x).

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Chapter 6: Factorisation of Polynomials - Exercise 6.5 [Page 33]

APPEARS IN

RD Sharma Mathematics [English] Class 9
Chapter 6 Factorisation of Polynomials
Exercise 6.5 | Q 16 | Page 33
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