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# Simplifying rational expressions: grouping

Sal simplifies and states the domain of (2x^2+13x+20)/(2x^2+17x+30). Created by Sal Khan.

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• 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? •  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?

• 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. • 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.
• Ok wait I am still very confused. In the video Sals factoring by grouping what is that? I need some further explanation • At when he had two of the (x + 4), why did he only write one for the final equation? Wouldn't it be (x + 4) ^2 (2x + 5)? • 2x(x+4)+5(x+4), let's write it like this: ac+bc. And a is 2x, b is 5, and since the two (x+4) are the same, let's call them both c.

Imagine that you have 4 apples, and your friend gives you 3 apples, you now have a total of 7 apples. That's the same here.

You have "a" much of "c" and "b" much of "c", and totally (since you need to add them) you have (a+b) much c.

And go back to the formula, you have "2x" much "(x+4)" and "5" much "(x+4)", And totally you have "(2x+5)" much of "(x+4)".

And "much = multiply", you got (2x+5)(x+4).

Hope that helps!
(1 vote)
• I don't understand why at and 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? • I don't understand how this first expression isn't a quadratic? • 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? • 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.
• when u factor the denominator wouldn't 2 and 15 work as a and b • 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? • 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. 