Therefore, you need:
Can you solve for that system of equations? If so, then the factorization comes out to be:
Pretty much, whenever you are given an equation, you have a set of methods to use:
1. Quadratic equation:
Then you just have to find an a and a b that are suitable to solve the two equations.
3. Completing the square:
If you can't factor an equation intuitively and you don't want to solve for the quadratic equation, then solve using this method:
The second problem will work for this one:
Step 1: Move the c term over.
Step 2: Divide the entire equation by what the is being multiplied by.
Step 3: Take half of the coefficient of the x-term and square it.
Step 4: Add this to both sides of the equation:
Step 5: factor left-hand side.
Step 6: Take the square root of both sides:
Step 7: Simplify right-hand side.
Step 8: Solve
Multiply the coefficient of the leading term and the constant term. We have 1 times 9 = 9 in this case.
Now you want to break 9 down into all possible pairs of factors:
Of these which pair adds up to the coefficient of the linear term, in this case -6? The -3, -3.
So -6 = -3 + -3. Rewrite your original problem using this:
<-- Now factor each term:
<-- Factor the common x - 3:
Try this process on your second problem and post the results.
What do you mean by the second part?
Say you have an equation:
I'm going to go over each way you can use to possibly solve this:
If the term of an equation is JUST , then the factors of the equation MUST have the following format:
There is no way around it, ok. I'll tell you my thoughts as I try to solve it intuitively.
Ok, well, if we use the FOIL method (First, Outer, Inner, Last) to solve the factors above we get:
First (First term of each factor):
Outer (Farthest left term of first factor, farthest right term of second factor):
Inner (Farthest right of first factor, farthest left term of second):
Last (Last term of each factor):
Which gives us the following:
Now, we can leave the out because it doesn't matter, so we have:
Now, we know that there is a 1 before the x in the original equation, so:
We also know that there was a -2 for the last term of the equation:
Now we need an a and a b that satisfies both equations, say we make a -1:
Now we need to check it using the second one:
It checks out, so, and
It's just guess and check.
Most of the time you can solve it this way, if not, then use the quadratic equation until you learn a new method from a tutor or instructor.