Question

how to factor:
1. x^3+8x^2+16x=0
2. x^3+x^2-25x=25

Answer 1

Ok the first one is pretty straight forward, if we factor an \(\large x\) out of each term, we'll be left with a quadratic in the brackets.\[\large x(x^2+8x+16)=0\]Understand that step?

Answer 2

yes

Answer 3

So from here, we want numbers that `multiply` to give us \(\large 16\), and those same numbers when we `add` them should give us \(\large 8\).
Those are the two numbers that will give us our factors.

\(\large 16\) and \(\large 1\).
\(\large 16 \times 1=16\)
But when we add them \(\large 16+1\neq8\).

So those values won't work out. Let's try some other multiples of 16.
How about \(\large 4\) and \(\large 4\), will those work?

Answer 4

yes

Answer 5

\(\large 4\times4=16\) and \(\large 4+4=8\)
Cool! So we can write our factors as two binomials with x and 4.
\[\large x(x^2+8x+16)=0 \qquad \rightarrow \qquad x(x+4)(x+4)=0\]
Good? :)

Answer 6

yes
so then i set both binomials equal to zero?

Answer 7

If the problem is asking you to find the `roots` or `solutions` then yes, you would use the `Zero Factor Property` ~ Set each factor equal to 0, and solve for x separately.\[\large x=0 \qquad \qquad x+4=0 \qquad \qquad x+4=0\]

But if the directions just say "factor this", then we were done with our last step.

Answer 8

ok thx for the help

Answer 9

np c:

Answer 10

ya math is my worst subject thats why i always need help with my math homework

Answer 11

what about the 2nd problem?

Answer 12

Ah that stinks c: we all have different strengths!

Hmm this next problem I'm a little confused on... I always forget how to factor cube functions... thinkinggg c:

Answer 13

ya it does since i want to go into medical school and have to major in it and its ok

Answer 14

Oh for this next one I think we can solve by grouping. Lemme try real quick before I explain it.

Answer 15

ok sure

Answer 16

\[\large x^3+x^2-25x=25\]Ok let's start by subtracting 25 from each side,\[\large \color{royalblue}{x^3+x^2}-25x-25=0\]From here, we'll factor out an x^2 from each of the blue terms.
\[\large x^2(\color{royalblue}{x+1})-25x-25=0\]

Answer 17

\[\large x^2(x+1)\color{orangered}{-25x-25}=0\]Then let's factor a -25 out of each orange term.\[\large x^2(x+1)-25(\color{orangered}{x+1})=0\]

Answer 18

Understand those steps? Try to digest that a moment c: Lemme know if you're confused so far.

Answer 19

i understand so far

Answer 20

This next step is a littleeeee tricky. We have 2 terms. They share a factor. Though, it's not a number or simply x^2 that we're factoring out this time.

We want to factor out \(\large (x+1)\) from each term.\[\large \color{#F35633}{x^2}(x+1)\color{#F35633}{-25}(x+1)=0 \qquad \rightarrow \qquad (x+1)\left(\color{#F35633}{x^2-25}\right)=0\]

Hopefully the color coding helps you see where everything went.

Answer 21

We have one more simplification we can do. I'll wait and see if you understand that part first though.

Answer 22

i understand it

Answer 23

For the last step, we'll need to recall this identity.
\(\large a^2-b^2 \qquad = \qquad (a-b)(a+b)\)
It tells us how to factor the `difference of squares`.

In our problem we have,\[\large (x+1)\left(x^2-25\right)=0\]Which we can write as,\[\large (x+1)\left(x^2-5^2\right)=0\]
Because we recognize that \(\large 25=5^2\).
Now we have the difference of squares in our right set of brackets. Can you see how it will factor?

Answer 24

(x-5)(x+5)?

Answer 25

Yes very good :)

Answer 26

Which gives us a final answer of,
\[\large (x+1)(x+5)(x-5)=0\]

Yay team!

Answer 27

yay thanks for the help @zepdrix