Question 1

I still don't get this video, I understood the first one but the second one is confusing me. What is Sal doing? If he is doing (2xsq + 13x + 20)/(2xsq + 17x + 30) Wouldn't that just be  
4x+10?

Accepted Answer

Dylan,

You can't just divide each term in the numerator by a corresponding term in the denominator, because each term in the numerator is being divided by the WHOLE denominator.

For example, the expression you provided (2x^2 +13x +20)/(2x^2+17x +30) is the same thing as 2x^2 / (2x^2+17x+30)  + 13x/(2x^2+17x+30) + 20/2x^2+17x+30)

Instead, to simplify the expression you could try factoring the top and the bottom of the expression.  This would be sort of like reducing a fraction like 45/90 into 1/2, except done with a more complicated expression:

For example, in the equation you listed the numerator can be factored into: 
(2x^2+13x+20)= (2x+5)(x+4)

And the denominator can be factored into: 
(2x^2+17x+30)=(2x+5)(x+6)

So that means the whole expression can be rewritten as:
(2x+5)(x+4)
-------------
(2x+5)(x+6)

Now, since there is a 2x+5 in the numerator divided by a 2x+5 in the denominator, the whole thing can be reduced to:

(x+4)/(x+6)

The only caveat is that the 2x+5 cannot be equal to 0, because then you would have a 0 in the denominator and the expression would be undefined.  That means that x cannot equal -2.5.  So, the best way to write the final answer would be:

(x+4)/(x+6)  with x not equal to -2.5

Does that help it make sense?

Question 2

1:00

Could someone tell me which video explains ( not how to use it )
the a*b, a+b  and why it works ?

I know I have seen it but I still don't quite understand
I can derive the a+b, a*b for a normalized 2nd degree equation.
But not for any other.
I still miss the explanation why it works out and which coefficients
enter the equation.

Accepted Answer

Suppose we have the following polynomial: x^2 + px + q, whose roots are x = a and x = b. We therefore know that it is equal to (x - a)(x - b). Expanding that gives us x^2 - ax - bx + ab = x^2 -(a+b)x + ab. So we know that p = -(a+b) and q = ab.

It also works when the coefficient of x^2 isn't 1. Suppose we have a polynomial px^2 + qx + r, whose roots are x = a and x = b. We therefore know that it's equal to p(x - a)(x - b) = p(x^2 - ax - bx + ab) = p(x^2 -(a+b)x + ab).
Set the two to be equal:
px^2 + qx + r = p(x^2 -(a+b)x + ab)
Divide by p:
x^2 + (q/p)x + r/p = x^2 -(a+b)x + ab
So we know that q/p = -(a+b) and r/p = ab.

Obviously, if a+b and ab are some ugly rationals, using this method is very impractical, but it works.

Question 3

Ok wait I am still very confused. In the  video Sals factoring by grouping what is that? I need some further explanation

Accepted Answer

These videos by Sal explain how to do grouping with polynomials: https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-quadratics-2/v/factor-by-grouping-and-factoring-completely Hope this helps!

Question 4

I don't understand why at 1:13 and 3:26 he splits up the 13x and 7x. Couldn't you just factor the problem without splitting the coefficients?  Do we have to do that for every factoring problem?

Accepted Answer

Did you actually factor this without grouping? If you can, great! If you're like me and grouping is faster, then use grouping.

Question 5

I don't understand how this first expression isn't a quadratic?

Accepted Answer

quadratics don't have x in the denominator

Question 6

I've been doing some practice on this and I usually get them wrong. The last one I did before giving up and coming here said the answer was x^2-2/x^2+2. I actually got that, but I kept cancelling out to get -1 (cancel out x^2 and get -2/2, reduce to -1). How do you know when to stop reducing? is there always supposed to be an x on both the numerator and the denominator?

Accepted Answer

The thing to remember with cancelling is that we can only cancel factors (things being multiplied).  In your example above, you cancelled terms (things being added/subtracted).  We stop cancelling when there are no more common factors between the numerator & denominator.
Since the binomials in: (x^2-2)/(x^2+2) are not factorable and they do not match (aren't common factors), we can't reduce the fraction any further.  
Hope this helps.

Question 7

when u factor the denominator wouldn't 2 and 15 work as a and b

Accepted Answer

No, because:
For Ax^2 +Bx +C
a*b = AC  
a + b = B

a+b has to be equal to 17 and 2+15 is 17 so that would work

However a*b has to be equal to AC = 60 and 2*15 is not equal to 60.

Question 8

how can this same solving method be used when one of the signs is minus such as the first one. The reason i ask is because on one of the little practice quizzes there is a problem :4y^2​​ −18y+18/​y^3​​ −6y^2​​ +9y. This problem is making my brain hurt... its not the material its just this problem is really throwing me off. Thank you . Also what would be the "numbers" that make this problem undefined?

Accepted Answer

First of all, move the common factors outside of the parentheses.
2 ( 2y^2 - 9y + 9 ) / [ y ( y^2 - 6y + 9 ) ]
Then factor both trinomials.
2 ( 2y - 3 ) ( y - 3 ) / [ y ( y - 3 ) ( y - 3  ] )
You can at this point easily notice the values that make the denominator zero ( 0 and 3 ), so those are the unallowed values of y that make the problem undefined.
Next, divide out the common factor from both numerator and denominator to get
2 ( 2y - 3 ) / [ y ( y - 3) ]  as the final answer.      
Hope this will help your brain relax!

Integrated math 3

Course: Integrated math 3 > Unit 13

Simplifying rational expressions: grouping

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