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=(-5+√(69) )/(2)` and
`x=(-5-√(69) )/(2)`

Answer 2

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

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) ((2x^(5))/(2x^(3)) + (10x^(4) )/(2x^(3)) -(22x^(3))/(2x^(3)))`

Divide expression in bracket

`(2x^(5))/(2x^(3)) = (2x^(5-3))/(2)`

`(2x^(5))/(2x^(3)) ={x^(2)}`

`(10x^(4) )/(2x^(3)) = (10x^(4-3) )/(2)`

`(10x^(4) )/(2x^(3)) = (10x)/(2)`

`(10x^(4) )/(2x^(3)) = {5x}`

`(22x^(3))/(2x^(3)) = (22x^(3-3))/(2)`

`(22x^(3))/(2x^(3)) = (22x^(0))/(2)`

`(22x^(3))/(2x^(3)) = (22 * 1)/(2)`

`(22x^(3))/(2x^(3)) = 11`

Bringing these results together, we have

`2x^(3) ((2x^(5))/(2x^(3)) + (10x^(4) )/(2x^(3)) -(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})`