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## Integrated math 2

### Course: Integrated math 2 > Unit 3

Lesson 5: Solving quadratics by factoring- Solving quadratics by factoring
- Solving quadratics by factoring
- Quadratics by factoring (intro)
- Solving quadratics by factoring: leading coefficient ≠ 1
- Quadratics by factoring
- Solving quadratics using structure
- Solve equations using structure
- Quadratic equations word problem: triangle dimensions
- Quadratic equations word problem: box dimensions
- Solving quadratics by factoring review

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# Solving quadratics by factoring: leading coefficient ≠ 1

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(155 votes)- 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.(42 votes)

- what does coefficient mean again?(15 votes)
- 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.(35 votes)

- At3:37to the end, how come you didn't add the 6 with your solution to make it look like 6(x+-10)(x+-10)?(10 votes)
- 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.(12 votes)

- 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(5 votes)
- 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.(15 votes)

- how would we factor out the expression y=3x^2+12-15 ?(3 votes)
- 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.(7 votes)

- 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(3 votes)
- i got same result by only regrouping (factoring 6 from the left side of the equation) is it correct ?(4 votes)
- 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)

- i am still confused on what a quadratic is, what is it?(2 votes)
- 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.(5 votes)

- Is there a way to determine if a quadratic equation can't be factored without using the quadratic formula?(3 votes)
- You can complete the square also, if you end up with a negative number opposite square or a non-perfect square number, it cannot be factored with out radicals or imaginary numbers.(3 votes)

- what if you have x² + 6x + 8 = 0(1 vote)
- Find two numbers which will multiply to 8 and add to 6.(5 votes)

## Video transcript

- [Voiceover] We have six
x squared, minus 120 x, plus 600, equals zero. Like always, pause this video, and see if you can solve for x, if you could find the x values
that satisfy this equation. Alright, let's work through this together. The numbers here don't seem
like outlandish numbers. They seem like something that
I might be able to deal with, and I might be able to factor,
so let's try to do that. The first thing I like
to do is see if I can get a coefficient of one, on
the second degree term, on the x-squared term. It looks like actually all of these terms are divisible by six. So if we divide both sides
of this equation by six, I'm still going to have
nice integer coefficients. Let's do that. Let's divide both sides by six. If we divide the left side
by six, divide by six, divide by six, divide by six. And I divide the right side by six. If I do that, and clearly
if I do the same thing to both sides of the equation, then the equality still holds. On the left-hand side, I am
going to be left with x squared, and then negative 120, divided by six. That is, let's see. 120 divided by six is 20. So that's minus 20 x. Then 600 divided by six, is 100. So plus 100, is equal
to zero divided by six. Is equal to zero. So let's see if we can factor, if we can express this quadratic as a product of two expressions. The way we think about this, and we've done it multiple times, if we have something, if we have x plus a, times x plus b, and this is
hopefully review for you. If you multiply that out,
that is going to be equal to, that equals to x squared, plus a, plus b, x, plus a b. What we want to do is
see if we can factor this into an x plus a, and an x plus b. A plus b, needs to be
equal to negative 20. That needs to be a plus b. And then a times b, right over here, that needs to be equal
to the constant term. That needs to be a times
b, right over there. Can we think of two numbers,
that if we take their product, we get positive 100, and
if we take their sum, we get negative 20? Well since their product is positive, we know that they have the same sign. They're both going to have the same sign. They're either both going to be positive, or their both going to be negative, since we know that we
have a positive product. Since their sum is negative, well they both must both be negative. You can't add up two positive
numbers, and get a negative. So they both must be negative. Let's think about it a little bit. What negative numbers,
when I add them together I get negative 20, when
I multiply it, I get 100? Well you could try to factor 100. You could say, well negative
two times negative 50, or negative four times negative 25. But the one that might jump out at you is this is negative 10,
times, I'll write it this way, negative 10, times negative 10, and this is negative 10, plus negative 10. So in that case, both our a and
our b, would be negative 10. And so we can rewrite the
left side of this equation as, we can rewrite it as, x, and I'll write it this way at first, x plus negative 10, times, x plus negative 10 again. X plus negative 10, and that is going to be equal to zero. So all I've done is I've
factored this quadratic. Another way, these are both
the same thing as x minus 10. I could rewrite this
as x minus 10, squared, is equal to zero. The only way that the left-hand side is going to be equal to zero, is if x minus 10 is equal to zero. You could think of this as taking the square root of both sides. It doesn't matter if
you're taking the positive or negative square root, or both of them. The square root of zero, is zero. So, we would say, that x minus
10 needs to be equal to zero. So x , adding 10 to both sides of this, you have x is equal to 10, is the solution to this quadratic equation, up here.