Question

17. The factors of x3-2x2-x+2 are
(A) (x-1)(x-1)(x-2)
(B) (x+1)(x+1)(x+2)
(C) (x+1)(x-1)(x-2)
(D) (x+1)(x+1)(x-2)​

Answer 1

first do trail and error method to get one factor

takex=1

then,

(1)³-2(1)²-1+2=0

so, x-1 is the factor of x³-2x²-x+2

now divide x³-2x²-x+2 by x-1

you get,

x²-x-2

Now factorise x²-x-2

you get,

x²-2x+x-2

=x(x-2)+1(x-2)

so, the factors of x³-2x²-x+2 is (x-1),(x-2),(x+1)

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Answer 2

$\purple{\bf{\underline{Answer:-}}}$

✅The correct option is (C) (x+1)(x-1)(x-2) ✔

$\pink{\bf{\underline{Step-by-step\: Explaination:-}}}$

$f(x) = {x}^{3} - 2 {x}^{2} - x + 2 \\ ⟹ f(1) = {1}^{3} - 2 \times {1}^{2} - 1 + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1 - 2 - 1 + 2 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0 \\ \\ ∴x = 1 \\ ⟹ x - 1 = 0 \\ \\ hence , \: (x - 1) \: is \: a \: product \: of \: f(x) \\ \\ f(x) = {x}^{3} - 2 {x}^{2} - x + 2 \\ \: \: \: \: = {x}^{3} - {x}^{2} - {x}^{2} + x - 2x + 2 \\\: \: \: \: = {x}^{2} (x - 1) - x(x - 1) - 2(x - 1) \\\: \: \: \: = (x - 1)( {x}^{2} - x - 2) \\ = (x - 1)(x + 1)(x - 2)$

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