Factorise p(x)=2x^3+3x^2+3x+2?

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Answer 1 · 2018-04-20T12:59:35

$p (x) = (x + 1) (2 x^{2} + x + 2)$ $p(x)=(x+1)(2x^2+x+2)$

$p (x) = 2 x^{3} + 3 x^{2} + 3 x + 2$ $p(x)=2x^3+3x^2+3x+2$

Rearrange

$p (x) = 2 x^{3} + 2 + 3 x^{2} + 3 x$ $p(x)=2x^3+2+3x^2+3x$

$p (x) = 2 (x^{3} + 1) + 3 (x^{2} + x)$ $p(x)=2(x^3+1)+3(x^2+x)$

$p (x) = 2 (x + 1) (x^{2} - x + 1) + 3 x (x + 1)$ $p(x)=2(x+1)(x^2-x+1)+3x(x+1)$

Take $x + 1$ $x+1$ as a common factor

$p (x) = (x + 1) (2 (x^{2} - x + 1) + 3 x)$ $p(x)=(x+1)(2(x^2-x+1)+3x)$

Simplify

$p (x) = (x + 1) (2 x^{2} + x + 2)$ $p(x)=(x+1)(2x^2+x+2)$