Question

Factorise: (x²-2x)²-11(x²-2x)+24​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

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$(x-4)(x+2)(x-3)(x+1) \ are \ factors \ of \ (x^2-2x)^2-11(x^2-2x)+24$

Step-by-step explanation:

$Given \ (x^2-2x)^2-11(x^2-2x)+24\\\\we \ have \ to \ factorise \ it.\\\\Let \ x^2-2x=y\\\\Now \ we \ get\\\\y^2-11y+24\\\\first \ factorise \ it$

$y^2-11y+24\\\\by \ splitting \ mid \ term\\\\we \ have \ to \ make \ equal \ (ac)=(+-b) \ as \ we \ need\\\\24\times1=8\times3 \ (ac)\\\\8+3=11 \ (+b)\\\\y^2-11y+24\\\\y^2-8y-3+24\\\\y(y-8)-3(y-8)\\\\(y-8)(y-3)\\\\y=3 \ or \ y=8$

$putting \ y=8 \ in \ x^2-2x=y\\\\x^2-2x=8\\\\x^2-2x-8=0\\\\x^2-4x+2x-8=0\\\\x(x-4)+2(x-4)=0\\\\(x-4)(x+2)=0\\\\now \ putting \ y=3 \ in \ x^2-2x=y$

$x^2-2x=3\\\\ x^2-2x-3=0\\\\x^2-3x+x-3=0\\\\x(x-3)+(x-3)=0\\\\(x-3)(x+1)=0$

$So, \ we \ factorise \ (x^2-2x)^2-11(x^2-2x)+24\\\\Thus, \ we \ get \ factor \ (x-4)(x+2)(x-3)(x+1)$