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## Algebra 1 (Eureka Math/EngageNY)

### Course: Algebra 1 (Eureka Math/EngageNY)>Unit 4

Lesson 6: Topic A: Lesson 5: The zero product property

Sal solves 6x²-120x+600=0 by first dividing by 6 and then factoring.

## Want to join the conversation?

• What do you do if you have a leading coefficient that is not 1 and you can't simplify all of the numbers like he did in the video?
For example:
5x^2+16x+12=0
• You simply multiply the coefficient of x^2 by the constant. Then ab = constant * coefficient of x^2. While a + b = well the coefficient of x.
So 5x^2+16x+12=0 . 12 * 5 = 60 so ab = 60. While a + b = 16
5x^2 + 10x + 6x + 12.........It's easy from here.
• what does coefficient mean again?
• The simplest definition of a coefficient is that it is the number in front of the variable(s) in a term.
For example, consider the polynomial: 5x^2 + 3x - 8
The 5 is the coefficient for the term 5x^2.
The 3 is the coefficient for the term 3x.
Hope this helps.
• At to the end, how come you didn't add the 6 with your solution to make it look like 6(x+-10)(x+-10)?
• Since you are finding solutions, not the equation, the 6 does not have any meaning because as Sal did in the beginning, 0/6 = 0. If you were trying to factor it as an equation, then you are correct in that f(x) = 6(x-10)(x-10) or f(x) = 6 (x-10)^2. This shows the whole quadratic function, not just the doubled up solution.
A solution is when f(x) = 0.
• so say i had the problem h^3+4h^2-12h-672 i can find what you factor out of the first two terms but i can't figure out what to take out to the second two terms to match the part in the parenthesizes you take out an h^2 from the first two and you left with (h+4) but i can't figure out the second part
• This looks like an example of factoring by grouping. The tip-off is the 4 terms and the leading exponent of 3.
h³ + 4h² - 12h - 672
so if you factor out the h², you get (h + 4) as you said.
After that, there is `a problem` with this method in this example. (h + 4) is not actually a factor of this polynomial, but it would have to be in order for there to be a way for us to find it again in the second set of terms. So, possibly you wrote the example down wrong? Or if you made it up, you would have to have something like - 48 as the final term. Then it would be h³ + 4h² - 12h - 48
Then you could follow this method:
(h³ + 4h²) - (12h + 48)
h²(h + 4) - 12 (h + 4)
So the factors would be (h + 4) and (h²- 12) and the roots would be -4, +2sqrt(3) and -2sqrt(3)
Now, with your `h³ + 4h² - 12h - 672`, if you graph this polynomial, there seem to be one positive root and two imaginary roots -- the positive root is 8
That means there is a factor of (h - 8), leaving h² + 12h + 84 as the other (quadratic) factor
This looks factorable, but is not. It has the two imaginary roots
h = - 6 + 4sqrt3 i
h = - 6 - 4sqrt3 i
So don't feel bad that you couldn't factor by grouping--this isn't a good victim for that method.
• how would we factor out the expression y=3x^2+12-15 ?
• First - Did you mean: y=3x^2+12x-15, with an x on the 12?
If yes, start factoring by factoring out a GCF=3
y=3(x^2+4x-5)
Then, factor the trinomial by finding 2 factors of -5 that add to 4. See if you can finish the factoring.
Comment back if you get stuck or you want to check your result.

Remember, you can always check your factors using multiplication to see if they create the original polynomial.
• i got same result by only regrouping (factoring 6 from the left side of the equation) is it correct ?
• I, actually did the same thing. But I remembered what you do to the right, you must do to the left, you can't factor 6 from 0 so, you where incorrect.
(1 vote)
• What about something like 9m^2-60m+100, where you cannot factor out a GCF? I know there's a way to solve it, but it's complicated and my brain isn't working and I was wondering if there was a video around here that deals with those specific types of problems
• i am still confused on what a quadratic is, what is it?
• A quadratic is a polynomial that can be written in the form: Ax^2 + Bx + C, where A, B and C are constants and A can not = 0. Here are examples of quadratics.
x^2 + 8x + 15
2x^2 - 32
Hope this helps.