Question

Can someone be kind and help me please and thank you....

What is the factored form for
9x^4-6x^3+18x^2-12x
4x^3+3x^2+8x+6
20p^3+40p^2+15p+30

Answer 1

for eachone?

Answer 2

yes please

Answer 3

ok

Answer 4

first one: 3x(x^2+2)(3x-2)

Answer 5

second one: (x^2+2)(4x+3)

Answer 6

@jjuden
Can you explain them too?

Answer 7

wondering what part of that is meant to be "help"

Answer 8

hold on

Answer 9

\(\bf 9x^4-6x^3+18x^2-12x\implies (9x^4-6x^3)+(18x^2-12x)\)

take common factors on each, see what you get

Answer 10

@jdoe0001 i know 3 is a common factor for each number but not sure how to do it

Answer 11

@jjuden thanks I'm pretty sure you were going to explain afterwards

Answer 12

well.. the 1st group has a common factor of \(\bf 3x^3\)
the 2nd group has a common factof or \(\bf 6x\)
try those

Answer 13

is there some type of formula to solve these equations

Answer 14

nope

Answer 15

you'd have to look at what's common to the terms

Answer 16

\(\bf 9x^4-6x^3+18x^2-12x\implies (9x^4-6x^3)+(18x^2-12x)
\\ \quad \\
(3\cdot {\color{brown}{ 3\cdot xxx}}x-{\color{brown}{ 3}}\cdot 2\cdot {\color{brown}{ xxx}})+({\color{brown}{ 6}}\cdot 3\cdot {\color{brown}{ x}}x-{\color{brown}{ 6}}\cdot 2{\color{brown}{ x}})\)

that should make them noticeable

Answer 17

\(\bf 9x^4-6x^3+18x^2-12x\implies (9x^4-6x^3)+(18x^2-12x)
\\ \quad \\
3x^3{\color{blue}{ (3x-2)}}+6x{\color{blue}{ (3x-2)}}\impliedby \textit{notice any other common factor?}\)

Answer 18

yes 3x and 2

Answer 19

well.. notice is a binomial though
thus \(\bf 9x^4-6x^3+18x^2-12x\implies (9x^4-6x^3)+(18x^2-12x)
\\ \quad \\
3x^3{\color{blue}{ (3x-2)}}+6x{\color{blue}{ (3x-2)}}\implies {\color{blue}{ (3x-2)}}(3x^3+6x)
\\ \quad \\
(3x-2)({\color{brown}{ 3x}}xx+{\color{brown}{ 3}}\cdot 2{\color{brown}{ x}})\impliedby \textit{see any other common factor?}\)

Answer 20

ok...i think i get it a little

Answer 21

I'm sorry math is really not my strong suit

Answer 22

well.. one of the binomials has a common factor
that is, 3x so
\(\bf 9x^4-6x^3+18x^2-12x\implies (9x^4-6x^3)+(18x^2-12x)
\\ \quad \\
3x^3{\color{blue}{ (3x-2)}}+6x{\color{blue}{ (3x-2)}}\implies {\color{blue}{ (3x-2)}}(3x^3+6x)
\\ \quad \\
(3x-2)({\color{brown}{ 3x}}xx+{\color{brown}{ 3}}\cdot 2{\color{brown}{ x}})\implies (3x-2)3x(x^2+2)\)

Answer 23

ohhhhhh ok ok ok...when you do it like that i understand better

Answer 24

the 2nd one is about the same thing
you group in 2 terms groups
so \(\bf 4x^3+3x^2+8x+6\implies (4x^3+3x^2)+(8x+6)\)

and get a common factor for each
you'd end up with a common binomial \(\bf 4x^3+3x^2+8x+6\implies (4x^3+3x^2)+(8x+6)
\\ \quad \\
x^2{\color{blue}{ (4x+3)}}+2{\color{blue}{ (4x+3)}}\implies {\color{blue}{ (4x+3)}}(x^2+2)\)

Answer 25

and the last one is about the same, \(\bf 20p^3+40p^2+15p+30\implies (20p^3+40p^2)+(15p+30)
\\ \quad \\
20p^2{\color{blue}{ (p+2)}}+15{\color{blue}{ (p+2)}}\)

Answer 26

so the common factor would be 20 and 15?

Answer 27

for the groups respectively, yes

Answer 28

if you multiply \(20p^2\) and distribute, you'd get the original 1st group

Answer 29

jesus take the wheel

Answer 30

5(4p^2+3)(p+2)???????

Answer 31

no wait sorry...5(2p^2+6)(p+4)

Answer 32

\(\bf 20p^3+40p^2+15p+30\implies (20p^3+40p^2)+(15p+30)
\\ \quad \\
20p^2{\color{blue}{ (p+2)}}+15{\color{blue}{ (p+2)}}\implies {\color{blue}{ (p+2)}}(20p^2+15)
\\ \quad \\
(p+2)(4\cdot 5pp+3\cdot 5)\implies (p+2)5(4pp+3)
\\ \quad \\
5(4p^2+3)(p+2)\)

Answer 33

dammit man

Answer 34

thanks sooo much for your help @jdoe0001

Answer 35

yw