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## Algebra (all content)

### Unit 10: Lesson 16

Factoring polynomials with quadratic forms- Factoring quadratics: common factor + grouping
- Factoring quadratics: negative common factor + grouping
- Factor polynomials: quadratic methods
- Factoring two-variable quadratics
- Factoring two-variable quadratics: rearranging
- Factoring two-variable quadratics: grouping
- Factor polynomials: quadratic methods (challenge)
- Factoring quadratics with common factor (old)

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# Factoring quadratics with common factor (old)

An old video where Sal factors 8k²-24k-144 by first taking out a common factor of 8 and then using the sum-product pattern. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- This may seem like a basic question, but how does he rewrite the equation as 8*(k+3)*(k-6)? don't you have to distribute the 8 onto both the k+3 and k-6 since he factored them out in the first place? It's not the same as the distributive property of multiplication? Help?(9 votes)
- If the '8' was distributed on to both the expression would look like this ...

(8k + 24)*(8k - 48)

and if you multiplied these two out you would get ...

(64k^2 - 384k + 192k - 1152) ----> (64k^2 -192k -1152)

which does not equal the original: 8k^2 - 24k -144(7 votes)

- Can't you factor this equation without simplifying the value of x^2?(4 votes)
- Yes, you can factor the trinomial without first factoring out any common factor. It may be much tougher to find the right combination of factors to make it work, though.

Also, in the end, for the equation to be factored completely, the factor will have to come out sooner or later. Here is the problem worked both ways:

8k^2 - 24k - 144= 0

8(k^2 - 3k - 18) = 0

8(k-6)(k+3) = 0

8k^2 - 24k - 144= 0

(8k + 24)(k - 6) = 0 *** This is tough to find the right factors.

8(k+3)(k-6)= 0(7 votes)

- If the equation were set equal to zero, couldn't you have just divided both sides by 8 and simply gotten the answer (k+3)(k-6)=0? I'm assuming that because it's an expression and not equation, we left the 8 in and didn't do algebraic operations. Am I correct?(4 votes)
- You can't just "set it equal to zero". Suppose k = 7

8(k+3)(k-6) = 8(7+3)(7-6) = 80

But (k+3)(k-6) = (7+3)(7-6) = 10

So, you cannot just ignore the constant except for those points where the rest of the expression equals zero (since any number times 0 equals 0).

Even if it were an equation, it might not be zero that it equaled.

Suppose 8(k+3)(k-6) = 176, dividing both sides by 8 leave (k+3)(k-6) = 22.

So, you must be VERY cautious about the constant in front of the other factors. You can only divide it out when the expression actually does equal zero (which is not the usual case).(6 votes)

- What is the difference between a trinomial and a polynomial?(2 votes)
- a trinomial is an expression with exactly three parts, and a polynomial is an expression with more than one.(8 votes)

- What happens if there are no 2 numbers compatible as the product of the last number, and sum of the second?(5 votes)
- Not all trinomial quadratics can be rationally factored. For example, it is not possible to rationally factor
`2x²+15x-3`

So, the answer is that you cannot factor such a trinomial. There is a test to determine whether it is possible to factor a trinomial quadratic expression. The test doesn't tell you what the factors are, but only whether there are any rational factors at all.

For a quadratic in the form of:`ax² + bx + c`

and a, b and c are all rational real numbers:

There can be rational factors if and only if`b²-4ac`

is either 0 or a perfect square.

For example, the above quadratic,`2x²+15x-3`

has`a = 2, b=15, c=−3`

Thus,`b²-4ac = 15² − 4(2)(−3) = 249`

This is not a perfect square or 0, thus it is impossible to rationally factor this polynomial.

Now let us test`10x²-51x-91`

(this is not easy to factor, so let us check whether it is even possible).`a = 10, b= -51, c = -91`

thus`b²-4ac = (-51)² - 4(10)(-91) = 2601 + 3640 = 6241`

←

√6241 = 79**this is a perfect square, so it can be factored**`(0 votes)

- HELP

What do you do for an equation that = 0 like

p^2 + 5p - 84 = 0(3 votes)- also if you can factor out something (2x^3+4x^2+2x can have 2x factored out), you can find another value.

from previous problem: 2x(x^2+2x+1) x=0 or ?

(x+1)(x+1) x=-1

therefore, x=0 or -1(2 votes)

- but how would foil something like (x-2)(x3)(x-5)(3 votes)
- Foil (x-2) and (x-5) then multiply it by (x^3)(3 votes)

- What happens when they aren't common factors?(3 votes)
- When there isn't a GCF...

If a=1, then find a 2 two numbers that multiply to C, and add to B.

EX:

x^2 + 11x + 24 8 & 3 works!

(x + 8)(x + 3) is your answer!

If a≠1, then you can group!

EX.

5x^2 - 17x + 6, -2x and -15x will replace B

(5x^2 - 2x)(-15x + 6)

x(5x-2)-3(5x-2)

(x-3)(5x-2) is your answer!

Hope this helped, there is probably a lesson on this.(1 vote)

- 2:40

how did you come up with +3 and -6?(1 vote)- Sal was looking for numbers that make -18 when multiplied together, such as 3 and -6.(5 votes)

- Sorry if this isn't supposed to be under this video,

But I have been searching Kahn Academy for a tutorial on factoring trinomials WITHOUT a common factor and that are NOT a perfect square. Such as:

2x^2 + 5x - 3 = 0

Could anyone lead me to a tutorial for this type of problem?

Thank you!(2 votes)- You can factor this by grouping.

1 Multiply 1st and last term (2*3=6)

2 Look at what the number is for the middle term (5)

3 What 2 numbers are a product of 6 and add up to 5?

6*-1=-6

6-1=5

4 Replace these 2 numbers for middle term (5) and attach an x to each number

5 Combine like terms (terms that you can be factored)

6 Then solve

2x^2+6x-1x-3

2x(x+3) -1(x+3)

(2x-1)(x+3)

I put the negative in front of the 1 so the numbers in the parentheses would be identical.

Is this answer correct?

Expand it out and if you get the original equation, then you have the correct answer.

(2x-1)(x+3)

2x^2+6x-1x-3

Combine like terms

2x^2+5x-3

https://www.khanacademy.org/math/algebra/multiplying-factoring-expression

Look at videos under section:

Factor by grouping(3 votes)

## Video transcript

Factor 8k squared
minus 24k minus 144. Now the first thing
we can do here, just eyeballing
each of these terms, if we want to simplify it a
good bit is all of these terms are divisible by 8. Clearly, 8k squared is divisible
by 8, 24 is divisible by 8, and 144-- it might not be as
obvious is divisible by 8-- but it looks like it is. 8 goes into one and 144,
8 goes into 14 one time. 1 times 8 is 8. Subtract, you get a 6. 14 minus 8 is 6. Bring down the 4. 8 goes into 64 eight times. So it goes into 144 18 times. So let's just factor
out an 8 of this. And then that will
simplify our expression. It will actually give us
a leading 1 coefficient. So this will become 8 times k
squared minus 24 divided by 8 is 3k minus 18. Now we have to factor
this business in here. And remember if anything has the
form x squared plus bx plus c, where you have a leading
1 coefficient-- this is implicitly a one--
we have that here in this expression
in parentheses. Then we literally
just need to-- and we can do this multiple
ways-- but we need to find two
numbers whose sum is equal to the coefficient on x. So two numbers whose sum
is equal to negative 3 and whose product is equal
to the constant term. And whose product is
equal to negative 18. So let's just think about the
factors of negative 18 here. Let's see if we can do
something interesting. So it could be 1. And since it's negative, one of
the numbers has to be positive, one has to be negative 1 and
18 is if it was positive. And then one of these could be
positive and then one of these could be negative. But no matter what
if this is negative and this is positive
then they add up to 17. If you switch them, then
they add up to negative 17. So those won't work. So either we could
write it this way, positive or negative 1, and
then negative or positive 18 to show that they have
to be different signs. So those don't work. Then you have positive
or negative 3. And then negative
or positive 6, just to know that they
are different signs. So if you have positive
3 and negative 6, they add up to
negative 3 which is what we need them to add up to. And clearly, positive 3 and
negative 6, their product is negative 18. So it works. So we're going to
go with positive 3 and negative 6 as
our two numbers. Now, for this example-- just
for the sake of this example-- We'll do this by grouping. So what we can do is we can
separate this middle term right here as the sum
of 3k negative 6k. So I could write the negative
3k as plus 3k minus 6k. And then let me
write the rest of it. So we have k squared up
here, plus 3k minus 6k, which is the same thing
as this over here. And then we have minus 18. And then all of that's
being multiplied by 8. Now we're ready to
group this thing. We can group these first two
terms, they're both divisible by k. And then we can group-- let
me put a positive sign-- let's group these
second two terms. So then we have 8 times--
I'll write brackets here instead of drawing
double parentheses. Brackets are really
just parentheses that look a little
bit more serious. Now let's factor out a k
from this term right here. I'm going to do this
in a different color. Let's factor out the k here. So this is k times k plus 3. And then we have plus. And then over here
it looks like we could factor out a negative 6. So let's factor
out-- I'm going to do this in a different color--
let's factor out a negative 6 over here. So plus negative
6 times k plus 3. So now it looks like we
can factor out a k plus 3. There's a k plus 3 times k, and
then we have a k plus 3 times the negative 6. So let's factor that out. So we have this 8 out
front, that's not changing. So let me write that
in the brackets. We're factoring out a k plus 3. So then we have the k plus
3 that we factored out. And then inside of that
we just have left this k. Instead of writing
plus negative 6, I could just write k minus 6. We factor out the k
plus 3 and we're done. And then we can rewrite this. The way we wrote it here
it's 8 times the product of k plus 3 times k minus 6. But we know from the
properties of multiplication, this is the exact same
thing as 8 times k plus 3 times k minus 6. And we are done.