Question

GCF and then factor

5x^5+10x^4-15x^3+10x^2

Answer 1

what number goes into each of the coefficients, 5, 10, 15, and 10?

Answer 2

5

Answer 3

and do any of the terms have variables in common?

Answer 4

no

Answer 5

@abhrag321 ?

Answer 6

\[\color{blue}{5x^5}+\color{green}{10x^4}-\color{red}{15x^3}+\color{orange}{10x^2}\]you're telling me that all the different colors do not at least once contain \(x\)?

Answer 7

They do.

Answer 8

my bad

Answer 9

no problem. every term contains at the minimum \(x^2\), correct?

Answer 10

yes

Answer 11

okay so the GCF is \(5x^2\)
pull out \(5x^2\) from each term of \(5x^5+10x^4-15x^3+10x^2\)

Answer 12

and that would be?

Answer 13

you're going to have input on this, I'm not here to do it for you, simply to explain and step you through the process.
divide each term by \(5\), then divide each term by \(x^2\)

Answer 14

1x^3 + 2x^2-3x+2x

Answer 15

i took a shot

Answer 16

good job, except the last term will no longer have an \(x\) since \(2x^2\div x^2=2\).
now you have \(5x^2(x^3+2x^2-3x+2)\), so to further factor, focus on what's inside the parenthesis.

Answer 17

now what?

Answer 18

\[x^3+2x^2-3x+2\]rearrange and group the terms:\[(x^3-3x)+(2x^2+2)\]pull out GCF's from each group:\[x(x^2-3)+2(x^2+1)\]
and that's about as much as it can be "broken down"

Answer 19

1 and 3/?

Answer 20

so your final answer is \(5x^2(x^3+2x^2-3x+2)\) or \(5x^2[x(x^2-3)+2(x^2+1)]\)

Answer 21

the term x^3+2x^2-3x+2
is not be factorized using integer value.
so I think the result will be after finding gcf ,
the factorization,
5x^2(x^3+2x^2-3x+2)

Answer 22

thank you. i get it a little more