Question

x^2+3x-40

Answer 1

what are you supposed to do?

Answer 2

factor?
graph?

Answer 3

actually, you can't graph that... so i guess it is factoring

Answer 4

is it?

Answer 5

factore the trinomial

Answer 6

yes its factoring

Answer 7

well that took a while

Answer 8

\[x ^{2}+3x-40\]
factors to

(x+8)(x-5)
because 8-5=3 (3x)
and 8(-5)=-40

Answer 9

alright thanks ;)

Answer 10

no prob

Answer 11

okay what about x^2+26x+120

Answer 12

(x+20)(x+6)
same reason as the last one.

Answer 13

alright thanks c:

Answer 14

@smored why couldn't you graph that?

to factor these trinomials:

\[(x+a)(x+b) = x^2 + ax + bx + ab = x^2 + (a+b)x + ab\]
To make that match our trinomial,

\[x^2 + (a+b)x + ab\]\[x^2 + 26x + 120\]Clearly \[(a+b)x = 26x\]\[a+b=26\]and\[ab = 120\]

So we look for a pair of factors of 120 that sum to 26. 6*20 is such a pair, giving us:
\[(x+6)(x+20) = x^2 + 6x + 20x + 6*20 = x^2 + 26x +120\checkmark \]

Answer 15

oh well is cause i really dont know how to do this because the teacher that i got didnt know how to teach good so i really didnt learch much in that class so right now im taking apex which is a computer class where yuhh make up a class that we faild so yea

Answer 16

learn*

Answer 17

lol what wbout 2x^2+7x+5

Answer 18

Okay, when the leading coefficient isn't 1, things get a bit more complicated. I use a technique called factoring by grouping. Here's how we do it:

Multiply the leading and trailing coefficients: here we have 2 and 5, so 2*5 = 10

Now find a pair of factors of 10 that add to the middle coefficient:

well, 2 and 5 :-)

Now write the polynomial again, except split the middle term with those coefficients:
\[2x^2+7x+5 = 2x^2 + 5x + 2x + 5\]Now group each pair of terms:\[(2x^2+5x) + (2x+5)\]Now factor each group:\[x(2x+5) + 1(2x+5)\]Notice that there is a common factor there of \((2x+5)\)? Factor it out:
\[(2x+5)(x+1) = 2x*x + 2x*1 + 5*x + 5*1 = 2x^2 + 2x+5x + 5 \]\[= 2x^2 + 7x + 5\checkmark\]

So \((2x+5)(x+1)\) is the factoring of \(2x^2 + 7x + 5\)

Answer 19

okay thank you ;)

Answer 20

Does that make sense? You can do it in the case where the leading coefficient is 1 as well.
For example:

\[x^2 + 26x + 120\]1*120 = 120, factors of 120 that add to 26 are 20 and 6, so
\[x^2 + 20x + 6x + 120\]\[(x^2+20x) + (6x+120) \]\[x(x+20) + 6(x+20)\]\[(x+20)(x+6)\]

Or maybe you're wondering if the order matters in the middle terms?

\[x^2 + 6x + 20x + 120\]\[(x^2+6x) + (20x+120)\]\[x(x+6)+20(x+6)\]\[(x+6)(x+20)\]Same answer.

Answer 21

ah alright i see now

Answer 22

It just takes some practice. You can always check your work by multiplying it out. If you don't get the same thing you started with, you made a mistake somewhere.

Answer 23

@whpalmer4 that was a great explanation, i would have been too lazy for that.

Answer 24

Once you've described the process a few dozen times, it doesn't take too much effort to do it again :-)

Answer 25

what about 2x^2+3x-14

Answer 26

Same steps:

\[2x^2+3x-14\]2*-14 = -28
We need a pair of factors of -28 that add to 3
-28*1
-14*2
-7*4
-4*7
-2*14
-1*28

-4 + 7 = 3

\[2x^2 - 4x + 7x - 14\]\[(2x^2-4x) + (7x-14)\]\[2x(x-2) + 7(x-2)\]Can you finish it?

Answer 27

is it (2x+7)(x-2)?

Answer 28

it is indeed!

Answer 29

yay! alright then

Answer 30

What about 14^2+29x-15?

Answer 31

A bit trickier. What is 14*-15? What pair of factors of that result add to 29?

Answer 32

oh never mind i already did it but thanks anyways

Answer 33

i dont know how to do this

(2a+3b)(2a-3b)
if says that i got to multiply the two binomials tp produce a diffrent of squares

Answer 34

to*

Answer 35

it*

Answer 36

Okay, let's do a a simplest case of difference of squares:
\[(a-b)(a+b) = a(a+b) - b(a+b) = a*a + a*b - b*a -b*b = a^2 -b^2\]

Answer 37

I should have done that with a different set of letters to avoid confusion:
\[(x-y)(x+y) = x(x+y) - y(x+y) = x*x + x*y - x*y -y*y = x^2 - y^2\]

So if you have a product of terms like you do, where one product term is thing1 - thing2, and the other product term is thing1 + thing2, you can skip immediately to the answer, which is (thing1)^2 - (thing2)^2

Answer 38

Here, we have \[(2a+3b)(2a-3b)\]"thing1" is \(2a\), "thing2" is \(3b\), so our answer is\[(2a)^2-(3b)^2 = 4a^2-9b^2\]

Answer 39

Same answer we would get if we went through all the effort:
\[(2a+3b)(2a-3b) = 2a*2a -2a*3b + 3b*2a -3b*3b = 4a^2 -6ab + 6ab - 9b^2 \]\[=4a^2-9b^2\]

Answer 40

ah i kinda understand it now so the answer is 4a^2-9b^2 right?

Answer 41

what about 25A^2B^4-16C^2D^2

Answer 42

\[25a^2b^4-16c^2d^2\]what is \[\sqrt{25a^2b^4}\](assuming that all the variables are >0)
What is \[\sqrt{16c^2d^2}\](again, assuming that all the variables are >0)

Answer 43

Both of those are squares: \(5ab^2*5ab^2=25a^2b^4\) and \(4cd*4cd = 16c^2d^2\)so we can factor that as\[(5ab^2-4cd)(5ab^2+4cd)\]
Checking our work by multiplying it out:
\[(5ab^2-4cd)(5ab^2+4cd) = 5ab^2*5ab^2+5ab^2*4cd-4cd*5ab^2-4cd*4cd\]\[=25a^2b^2+20ab^2cd-20ab^2cd-16c^2d^2 = 25a^2b^2-16c^2d^2\checkmark\]

Answer 44

ah i kinda get it now

what about x^2=?x-36

Answer 45

What's the ? doing in there?

Answer 46

wait hold up i did it wrong its this
x^2+?x-36

Answer 47

what are you supposed to do with it?

Answer 48

well it says that in the polynomial below, what number should replace "?" to produce a difference of squares?

Answer 49

okay, well, why didn't you say so in the first place?

Answer 50

does a difference of squares have a middle term?

Answer 51

i dont know

Answer 52

\[(a+b)(a-b) = a^2-b^2\]Is there anything that looks like it should be called a middle term there?

Answer 53

How many terms are there? How many terms in the problem given?

Answer 54

well im guessing that there"s two terms

Answer 55

that's right, \(a^2\) and \(-b^2\) are the terms in our "prototype" difference of squares. So how do we make \[x^2+?x -36 \]have only two terms, both of which are squares?

Answer 56

is it 7 and 36?

Answer 57

we want to make\[x^2+ax-36\] look like \[x^2-36\]what would be a value of \(a\) that would accomplish that?

Answer 58

Hint: what is 234235234.234235 * 0?

Answer 59

Another hint: what is 0*x

Answer 60

ummm i think its 0

Answer 61

right. so if you substitute 0 for the question mark and simplify, what do you get?

Answer 62

\[x^2 + 0x - 36= x^2 -36\]