Question

FACTORING QUADRATIC EXPRESSIONS:
x^2-16x+63

and


9p^2+73pr+70r

Answer 1

7 and 9 are factors of 63; and 7 plus 9 sum to 16 so it is just a matter of finding the correct signs
(x-7) (x-9)

Answer 2

I got that, but is that all that I do?

Answer 3

For x^2-16x+63, we can factor using a simple method I have found to work for myself.
However, it helps to know the following information first.
The standard form of a quadratic is ax^2+bx+c.

For your given problem, x^2-16x+63. a=1, b=-16 and c=63

So when we factor, we are trying to split the quadratic, or other polynomial, into the parts that can multiply together to create it!
This is similar to how we can factor 2-2x to look like 2(1-x).

To factor any quadratic, follow this method.
Step 1: Make two sets of open parentheses.
Ex: ( ) ( )
Step 2: Plug the ax term (not ax^2) as the first value for each pair of parentheses.
Ex: (1x+ ) (1x+ )
Step 3: Multiply the a term by the c term. Then, find two numbers that multiply together to give you your c term and add up to the b term.
Ex: a=1, c=63, so a*c=63. 7 and 9 are both factors that multiply together to 63 and add to 16. But, b=-16, so we will need both values to be negative.
Step 4: Plug those values in after the addition sign. This can be in any order.
Ex: (1+(-7)) (1x+(-9)) , so (1x-7)(1x-9)
Step 5: Factor out any GCF in either set of parentheses between the terms within if one exists and get rid of it altogether.
Ex: There isn't one in your problem but an example is (2x+6) would be 2(x+3). Be sure to remove the 2 altogether, I usually scribble it out.

Your answer here is (x-7)(x-9)=0
You can set each factor to zero to find what x is equal to if the problem asks.
x-7=0, x=7
x-9=0, x=9

I'm not sure about your second question just yet though.

Answer 4

9p^2+73pr+70r

This can't be factored. The only way this can be factored is if either (a) 70r is squared, or (b) if 9p^2 has an r term with it.

For the sake of argument lets say you in fact do have 9p^2r +73pr+70r, for instance.
First, factor out r.
r(9p^2+73p+70)
Now you can factor out the stuff inside the parentheses. However, if you factor out a GCF initially, you MUST keep it. If you follow the method I presented to you then factor out a GCF from within the resulting parentheses later on, but keep r and not your other "garbage" GCFs you factor out.

Answer 5

Thanks on the great notes....one more question.....so if it asks me to solve expressions that do not have matching terms, theres no other way they can be simplified?

Answer 6

No, not really. You'll be left with a mess of numbers and variables anyway you try to cut it.

Also, if you try to factor, and it turns out that you can't multiply a by c and get two values that multiply to that number AND add up to the b term, then your quadratic is determined unfactorable. There is another method to factor such an equation. Actually, two. Completing the square and the quadratic equation.

Answer 7

A good example I can find of this is the equation x^2+6x+1.
For this, you either have to complete the square or use the quadratic formula.
The quadratic formula is usually easiest and looks like this:
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
All you have to do for this is plug in the values for a, b, and c from your original equation, then solve.

Answer 8

Nevermind, haha. x^2+6x+1 isn't a good example because it has negative numbers but the concept is still the same.
x^2+3x+5 is a better example

Would you like for me to show you how to solve that with the quadratic formula?
Or would you like me to teach you completing the square too?
Or, are you tired yet?

Answer 9

imaginary* not negative

Answer 10

I'm a little tired, lol. But I've got the concept of it. Thank you so so much.

Answer 11

Ok, if you need any more help I would highly recommend either asking questions here or go to khan academy and watch their videos.

Answer 12

okay, thanks. if i need more help i'll surely come back here.