Trying to find the right answer to this..

Let f(x) = x 4 + x 3 - 7x 2 - x + 6 By trial, f(1) = 0 and f(2) = 0 So by factor theorem, (x - 1) and (x - 2) are factors of f(x). (x - 1) (x - 2) = x 2 - 3x + 2 Now, f(x) = x 2 (x 2 - 3x + 2)+ 4x (x 2 - 3x + 2) + 3 (x 2 - 3x + 2) = (x 2 - 3x + 2) (x 2 + 4x + 3) = (x - 1) (x - 2) (x + 1) (x + 3)

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Polynomials

Short Answer Type

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open parentheses straight x minus 1 half close parentheses

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130. Using factor theorem, factorise the polynomial x⁴ + x³ - 7x² - x + 6.

Let f(x) = x⁴ + x³ - 7x²- x + 6
By trial, f(1) = 0 and f(2) = 0
So by factor theorem, (x - 1) and (x - 2) are factors of f(x).
(x - 1) (x - 2) = x² - 3x + 2
Now, f(x) = x² (x² - 3x + 2)+ 4x (x² - 3x + 2) + 3 (x² - 3x + 2)
= (x² - 3x + 2) (x² + 4x + 3)
= (x - 1) (x - 2) (x + 1) (x + 3)