1. ## Factorising

hi, just wondering if anyone can give me any help with this maths equation:

X2-6x+9=0
6x2+x-15=0

i have tried to wrk this equation out but un-sure
thanks jenko!

2. Originally Posted by jenko
hi, just wondering if anyone can give me any help with this maths equation:

X2-6x+9=0
6x2+x-15=0

i have tried to wrk this equation out but un-sure
thanks jenko!
For the first one:

$(x+a)(x+b)$

Ignoring the $x^2$ term:

$ax + bx + ab$

$ax + bx = 6x$

$ab = 9$

$a = \frac{9}{b}$

$\frac{9}{b}x + bx = 6x$

$x(\frac{9}{b} + b) = 6x$

$\frac{9}{b} + b = 6$

$\frac{9+b^2}{b} = 6$

$9+b^2 = 6b$

$b^2 -6b + 9$

$(b-3)(b-3)$

b = 3

$ax + 3x = 6x$

$a + 3 = 6$

a = 3

Answer: $(x+3)(x+3)$

Someone else can do the second one...

3. anyone else have a easier way for me to understand this? no offence thanks very much

4. Originally Posted by jenko
anyone else have a easier way for me to understand this? no offence thanks very much
Basically, you need two numbers that when multiplying each other will equal nine. These same two numbers must also add up to negative six. This is for the first equality. You can use what you learn from this example for the second problem.

Therefore, you need:

$ab=9$
$a+b=-6$

Can you solve for that system of equations? If so, then the factorization comes out to be:

$(x+a)(x+b)$

5. Originally Posted by colby2152
Basically, you need two numbers that when multiplying each other will equal nine. These same two numbers must also add up to negative six. This is for the first equality. You can use what you learn from this example for the second problem.

Therefore, you need:

$ab=9$
$a+b=-6$

Can you solve for that system of equations? If so, then the factorization comes out to be:

$(x+a)(a+b)$
Correction: $(x+a)(x+b)$

Pretty much, whenever you are given an equation, you have a set of methods to use:

$x = \frac{-b \frac{+}{-}\sqrt{b^2-4ac}}{2a}$

2. Intuition:

Well, if:

$a + b = 6$

and

$ab = 9$

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:

$6x^2 + x - 15 = 0$

Step 1: Move the c term over.

$6x^2 + x = 15$

Step 2: Divide the entire equation by what the $x^2$ is being multiplied by.

$x^2 + \frac{1}{6}x = \frac{15}{6}$

Step 3: Take half of the coefficient of the x-term and square it.

$\frac{1}{6}(\frac{1}{2}) = \frac{1}{12}$

$(\frac{1}{12})^2 = \frac{1}{144}$

Step 4: Add this to both sides of the equation:

$x^2 + \frac{1}{6}x + \frac{1}{144} = \frac{15}{6} + \frac{1}{144}$

Step 5: factor left-hand side.

$(x+\frac{1}{12})^2 = \frac{361}{144}$

Step 6: Take the square root of both sides:

$(x + \frac{1}{12}) = \sqrt{\frac{361}{144}}$

Step 7: Simplify right-hand side.

$(x + \frac{1}{12}) = \frac{\sqrt{361}}{12}$

Step 8: Solve

$x = \frac{+}{-}(\frac{\sqrt{361}}{12} - \frac{1}{12})$

6. Aryth, thanks for fixing my typo!

7. how do i do the second part of the equation i have no idea how to do this?

8. Originally Posted by jenko
X2-6x+9=0
Let me give you a step-by-step.

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:
1, 9
3, 3
-1, -9
-3, -3

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:
$x^2 - 6x + 9$

$= x^2 + (-3 + -3)x + 9$

$= x^2 - 3x - 3x + 9$

$= (x^2 - 3x) + (-3x + 9)$ <-- Now factor each term:

$= x(x - 3) + -3(x - 3)$ <-- Factor the common x - 3:

$= (x - 3)(x - 3) = (x - 3)^2$

Try this process on your second problem and post the results.

-Dan

9. Originally Posted by jenko
how do i do the second part of the equation i have no idea how to do this?

What do you mean by the second part?

Say you have an equation:

$x^2 + x - 2 = 0$

I'm going to go over each way you can use to possibly solve this:

1. Intuition

If the $x^2$ term of an equation is JUST $x^2$, then the factors of the equation MUST have the following format:

$(x+a)(x+b)$

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):

$x*x = x^2$

Outer (Farthest left term of first factor, farthest right term of second factor):

$x*b = bx$

Inner (Farthest right of first factor, farthest left term of second):

$a*x = ax$

Last (Last term of each factor):

$a*b = ab$

Which gives us the following:

$x^2 + bx + ax + ab = 0$

Now, we can leave the $x^2$ out because it doesn't matter, so we have:

$bx + ax + ab = (b + a)x + ab$

Now, we know that there is a 1 before the x in the original equation, so:

$b + a = 1$

We also know that there was a -2 for the last term of the equation:

$ab = -2$

Now we need an a and a b that satisfies both equations, say we make a -1:

$b - 1 = 1$
$b = 2$

Now we need to check it using the second one:

$(-1)(2) = -2$

It checks out, so, $a = -1$ and $b = 2$

So:

$(x-1)(x+2)$

Therefore:

$x=1;x=-2$

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.

10. is this the final answer?

2x-3=0
3x+5=0