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Fully factorise: x³+3x²-6x-8​

User Dgatwood
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2 Answers

14 votes
14 votes

Answer:

(x-2) (x+1) (x+4)

Explanation:

x^3+3x^2-6x-8

Rearranging the terms,

(x^3 - 8) + (3x^2 - 6x)

a^3 - b^3 = (a-b) (a^2 + ab + b^2)

(x^3 - 8) = (x^3 - 2^3) = (x-2) (x^2 + 2x + 4)

(3x^2 - 6x) = 3x(x-2)

So, x^3 + 3x^2 -6x -8

= (x-2) (x^2 + 2x +4) + 3x(x-2)

(x-2) is the common factor,

= (x-2) ( x^2 +2x+4 + 3x)

= (x-2) (x^2 + 5x + 4)

= (x-2) (x^2 + 4x + x + 4)

= (x-2) (x(x+4) +1(x+4))

= (x-2) (x+1) (x+4)

User RNA
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2.8k points
15 votes
15 votes
I’m not sure if this is correct
Fully factorise: x³+3x²-6x-8​-example-1
User Qasta
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3.0k points