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### Course: Algebra (all content)>Unit 10

Lesson 16: Factoring polynomials with quadratic forms

# 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?
• 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
• Can't you factor this equation without simplifying the value of x^2?
• 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
• 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?
• 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).
• What is the difference between a trinomial and a polynomial?
• a trinomial is an expression with exactly three parts, and a polynomial is an expression with more than one.
• What happens if there are no 2 numbers compatible as the product of the last number, and sum of the second?
• 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`
• HELP
What do you do for an equation that = 0 like
p^2 + 5p - 84 = 0
• 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
• but how would foil something like (x-2)(x3)(x-5)
• Foil (x-2) and (x-5) then multiply it by (x^3)
• What happens when they aren't common factors?
• 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!

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)

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

• 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.
• 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!
• 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.
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