Question

Factor completely 2x5 + 10x4 − 22x3.

Answer 1

Given equation is :

$2x^{5}+10x^{4}-22x^{3}$=0

Taking out $2x^{3}$ common we get,

$x^{2}+5x-11$=0

Now solving this we get,

As 11 is not factor-able, we will solve it using the formula,

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$

a=1 b=5 and c=-11

we get,

$x=\frac{-5+\sqrt{69} }{2}$ and $x=\frac{-5-\sqrt{69} }{2}$

Answer 2

Answer:

The factor of $2x^{5} + 10x^{4} - 22x^{3}$ is $2x^{3} ({x^{2}}{} + {5x^{}}} - {11})$

Step-by-step explanation:

Given:

$2x^{5} + 10x^{4} - 22x^{3}$

Required:

Factorize

To factor a number means to take the number apart to find its factors

To factor $2x^{5} + 10x^{4} - 22x^{3}$ means we look for factors that can divide individual unit of the algebraic expression to the barest minimum

One factor that can divide through is $2x^{3}$

We then divide each individual unit of the expression with $2x^{3}$

So, we write $2x^{5} + 10x^{4} - 22x^{3}$ as

$2x^{3} (\frac{2x^{5}}{2x^{3}} + \frac{10x^{4} }{2x^{3}} -\frac{22x^{3}}{2x^{3}})$

Divide expression in bracket

$\frac{2x^{5}}{2x^{3}} = \frac{2x^{5-3}}{2}$

$\frac{2x^{5}}{2x^{3}} ={x^{2}}$

$\frac{10x^{4} }{2x^{3}} = \frac{10x^{4-3} }{2}$

$\frac{10x^{4} }{2x^{3}} = \frac{10x}{2}$

$\frac{10x^{4} }{2x^{3}} = {5x}$

$\frac{22x^{3}}{2x^{3}} = \frac{22x^{3-3}}{2}$

$\frac{22x^{3}}{2x^{3}} = \frac{22x^{0}}{2}$

$\frac{22x^{3}}{2x^{3}} = \frac{22 * 1}{2}$

$\frac{22x^{3}}{2x^{3}} = 11$

Bringing these results together, we have

$2x^{3} (\frac{2x^{5}}{2x^{3}} + \frac{10x^{4} }{2x^{3}} -\frac{22x^{3}}{2x^{3}})$ = $2x^{3} ({x^{2}}{} + {5x^{}}} - {11})$

Hence the factor of $2x^{5} + 10x^{4} - 22x^{3}$ is $2x^{3} ({x^{2}}{} + {5x^{}}} - {11})$