Question

how would I multiply (x+3)(x-3)(x+5)

Answer 1

Use the distributive property: \[
c(a+b) = ca+cb
\]

Answer 2

all you have to do is simplify...

Answer 3

\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)

\(\Huge \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\:\mathbb Welcome\:\:\mathbb To\:\:\mathbb Open\mathbb Study}\)
\(\large \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-\mathbb Snuggie\mathbb Lad }\)
\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)

Answer 4

For example:\[
(x+3)(x-3)(x+5) = (x+3)(x-3)x+(x+3)(x-3)5
\]

Answer 5

First: (x+3)(x-3) is of the form:
(a+b)(a-b)=a²-b², so that would be x²-9.

You could also FOIL:
(x+3)(x-3)
First: x²
Outer: x*-3=-3x
Inner: 3*x=3x
Last: 3*-3=-9

Add up: middle terms vanish. Result: x²-9.

But it is simpler to remember (a+b)(a-b)=a²-b² imo.

Answer 6

use the foil method to foil (x+3)(x-3) then foil that result with (x+5)
it should look something like this (x^2+9)(x+5) then you end up with x^3+5x^2+9x+45

Answer 7

Now you have (x²-9)(x+5).

Again: FOIL it to get the final result!

Answer 8

acutally it would be x^3+5x^2-9x-45

Answer 9

Use FOIL to multiply the first two binomials.
Then simplify that product.
Then multiply every term of the resulting product by every term of the remaining binomial.

Answer 10

FOIL (x²-9)(x+5):

First: x²*x=x³
Outer: x²*5=5x²
Inner: -9*x=-9x
Last: -9*5=-45

Put it together: x³+5x²-9x-45

Answer 11

@ZeHanz so it would be -4x^5-45?

Answer 12

No, you cannot put together terms with x³ and with x². They stay apart, because:

x³=x*x*x and x²=x*x

Answer 13

Only if you MULTIPLY, you get that:
\(2x^3 \cdot 3x^2=2\cdot x\cdot x\cdot x\cdot 3\cdot x\cdot x=2\cdot 3\cdot x\cdot x\cdot x\cdot x\cdot x=6x^5 \)

Answer 14

so would would it be once I added it?

Answer 15

The answer was given to you before you tried to combine unlike terms, kimmygin.

Answer 16

In the post right before you addressed ZeHanz.

Answer 17

so once I foiled x^2-9 and x+5 that's my polynomial??

Answer 18

That is correct.

Answer 19

are you sure?

Answer 20

Well, if I am not, then my students at work should be taking a different professor....lol

Answer 21

The answer after expanding is: x³+5x²-9x-45. You cannot go further!

Answer 22

oh gosh I wrote the problem wrong, the roots are (x+2)(x-3)(x+5) so could I just combine +2 and -3 and then foil that to x+5 itd be the same thing right

Answer 23

Well, because you now do not have (a+b)(a-b), things will be more difficult:
E.g. FOIL (x+2)(x-3), this becomes x²-x-6.

Now do this:

(x²-x-6)(x+5), so do 6 multiplications and clean up (be sure to add together only like terms!)

Answer 24

how the heckkkk do I multiply that

Answer 25

do I do distributive with all 3 numbers in the first parenthesis

Answer 26

@ZeHanz

Answer 27

Just take every number in the first parentheses and multiply wirh every number in the second. That is whay I said: 6 multiplications:

(x²-x-6)(x+5) =
x²*x = x³
x²*5 = 5x²
-x*x = -x²
-x*5 = -5x
-6*x = -6x
-6*5 = -30

Add up, and only take together the like terms!

Answer 28

x^3+4x^2-11x-30??

Answer 29

@ZeHanz

Answer 30

what if I was just suppoed to write it in factored form?

Answer 31

You answer is right.
For finding roots, the form (x+2)(x-3)(x+5) is much more useful:

(x+2)(x-3)(x+5) = 0 can be solved by just looking at the different factors and setting them 0:

x+2=0 --> x = -2
x-3=0 --> x = 3
x+5=0 --> x = -5

Done!

This is why we bother to factor a formula...

Answer 32

omg so I did all that work for nothing haha

Answer 33

Yeah, me too :D

Answer 34

im sorry lol

Answer 35

You are forgiven!