# How do you factor y= 2x^3 - 5x^2 + 4 ?

Feb 29, 2016

$y = \left(x - 2\right) \left(2 {x}^{2} - x - 2\right)$

#### Explanation:

y=2x^3-5x^2+4 = 2x^3-4x^2-x^2+4 = 2x^2(x-2) - (x^2-4) =2x^2(x-2) - ((x+2)(x-2)) = (x-2)(2x^2-(x+2) = (x-2)(2x^2-x-2)[Answer]

Aug 18, 2016

Let $g \left(x\right) = y = 2 {x}^{3} - 5 {x}^{2} + 4$.

We observe that sum of the co-effs.$= 2 - 5 + 4 = 1 \ne 0 , s o , \left(x - 1\right)$ is not a factor.

Also, the sum co-effs. of odd powered terms $= 2 \ne - 1 = - 5 + 4$ = the sum of co-effs. of even powered terms, so #(x+1) can not be a factor.

Now to guess the probable factors of $g \left(x\right)$, we look at its leading co-eff. $= 2 ,$ having factors $1 , 2$ and the const. term$= 4 ,$ having factors $1 , 2 , 4.$

The probable factors can be guessed by multiplying $x$ with the factors of leading co-eff. and taking with this, $\pm$ the factors of the const. term. This leads us to the guess of probable factors of $g \left(x\right)$ as $x \pm 1 , x \pm 2 , x \pm 4 , 2 x \pm 1 , 2 x \pm 2 , 2 x \pm 4.$

Among these, we have already verified the factors $x \pm 1 , 2 x \pm 2 = 2 \left(x \pm 1\right) .$

Next we verify whether $\left(x - 2\right)$ is a factor of $g \left(x\right)$ by checking $g \left(2\right) = 2 \cdot 8 - 5 \cdot 4 + 4 = 16 - 20 + 4 = 0 ,$, so, (x-2) is a factor.

Now we arrange the terms of $g \left(x\right)$ in such a way (shown below) that $\left(x - 2\right)$ turns out as a common factor :-

$g \left(x\right) = 2 {x}^{3} - 5 {x}^{2} + 4 = \underline{2 {x}^{3} - 4 {x}^{2}} - \underline{{x}^{2} + 2 x} - \underline{2 x + 4} = 2 {x}^{2} \left(x - 2\right) - x \left(x - 2\right) - 2 \left(x - 2\right) = \left(x - 2\right) \left(2 {x}^{2} - x - 2\right)$.

Enjoy maths.!