Question

Factonise the polynomial: (x²-x) ² - 8 [x^2-x]+12​

Answer 1

Answer:

$\left(x^2-x\right)^2-8\left(x^2-x\right)+12\\= \left(\left(x^2-x\right)^2-2\left(x^2-x\right)\right)+\left(-6\left(x^2-x\right)+12\right)\\= \left(x^2-x\right)\left(\left(x^2-x\right)-2\right)-6\left(\left(x^2-x\right)-2\right)\\= \left(\left(x^2-x\right)-2\right)\left(\left(x^2-x\right)-6\right)\\= \left(x^2-x-2\right)\left(x^2-x-6\right)\\= \left(x+1\right)\left(x-2\right)\left(x^2-x-6\right)\\= \left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x-3\right)$

MARK ME AS BRAINLIEST !!

Answer 2

Answer:

:

\begin{gathered}\left(x^2-x\right)^2-8\left(x^2-x\right)+12\\= \left(\left(x^2-x\right)^2-2\left(x^2-x\right)\right)+\left(-6\left(x^2-x\right)+12\right)\\= \left(x^2-x\right)\left(\left(x^2-x\right)-2\right)-6\left(\left(x^2-x\right)-2\right)\\= \left(\left(x^2-x\right)-2\right)\left(\left(x^2-x\right)-6\right)\\= \left(x^2-x-2\right)\left(x^2-x-6\right)\\= \left(x+1\right)\left(x-2\right)\left(x^2-x-6\right)\\= \left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x-3\right)\end{gathered}

(x

2

−x)

2

−8(x

2

−x)+12

=((x

2

−x)

2

−2(x

2

−x))+(−6(x

2

−x)+12)

=(x

2

−x)((x

2

−x)−2)−6((x

2

−x)−2)

=((x

2

−x)−2)((x

2

−x)−6)

=(x

2

−x−2)(x

2

−x−6)

=(x+1)(x−2)(x

2

−x−6)

=(x+1)(x−2)(x+2)(x−3)