Help with factoring an equation with variables and exponents...?

Question

z^2-8z+12

I'd love to know how to do this, not just an answer. I'm terrible at every type of math other than basic and geometry. I am required to pass a test next week on differenent levels so any help wouold be much appreciated. thanks

Answer 1

You have a quadratic equation whose factors are (z-2) and (z-6); that is, (z-2) * (z-6) = z^2-8*z+12.
How does one get this answer?
Well you have the z^2 term, which tells you there are two factors that we'll call z+a and z+b; now we have to find a and b. What do we know about a and b? We know that the product of z+a and z+b is z^2 + a*z + b*z + a*b, and this means that a*b=12 and a+b=-8. It's easy to see that a=-2 and b=-6 satisfies these equations, but there is a method that works for all cases (even when the answer is not easy to see) called "completing the square" (see ref.). Read the first example at that web page, and I'll list below the equations to substitute for each equation in the example:
z^2-8*z+12=0
z^2-8*z=-12 (eq. 2)
z^2+2*b*z+b^2=(z+b)^2 (binomial formula)
z^2-8*z+16=(z-4)^2 (binomial, completed square where b=-4)
z^2-8*z+16=-12+16=4 (adding 16 to both sides of eq. 2)
(z-4)^2=4 (from above 2 eqs)
z-4=+2 and z-4=-2
z=6 and z=2
z^2-8z+12=(z-6)*(z-2)
I hope this helps.

Answer 2

the same variable (z) can be factored out like this:
look at the highest exponent (z^2), that has to be the first in the factor, so:
(z +/- ?) ( z +/- ?)
next, look at the signs. The 12 is +, but the eight is -, the two signs both have to be -, because a + and +, muliplied out results in all +; a + and a -, will result in the lowest ranked exponent (12) being -, and the 8 could be either.

so now we have : (z - ?) ( z -?)
now what two numbers added together will equal 8, but multiplied will equal 12?
1 + 7 = 8, 1*7 = 7 wrong
2+6=8, 2*6 = 12, correct
so you have
(z-2)(z-6) = z^2 - 6z - 2z + 12 = z^2 - 8z + 12
that help?

Answer 3

z^2-8z+12

First, you're looking at how you need to factor this. z^2 = z * z

Therefore, your factors would begin as: (z ) (z )

Looking at the equation, you must determine what factors of the number 12 can also be either added or subtracted to equal 8. In this instance, 2 * 6 = 12. Notice that 6+2=8. Hence, the factored equation becomes: (z 2)(z 6)

To determine what the operators will be (+ or -) for the factored equation, you need to refer back to your original polynomial z^2-8z+12

In the polynomial, the 8 is a negative number (-8z) and 12 is a positive number (+12). Therefore, you have to set the operators in the factored equation for 6*2 to equal +12 (positive number) BUT ALSO where 6+2 equals -8 (negative number).

(-6) + (-2) = -8. (Two negative numbers added equals a negative number.)

Also, (-6) * (-2) = +12 (Two negative numbers mulitplied equals a positive number.)

Based on the facts, the operator in the factored equation will be:
(z - 2) (z - 6).

To check this, multiply them using the FOIL method

z *z = z^2
z * (-6) = -6z
(-2) * z = -2z
(-2) * (-6) = +12

You can combine like factors; which would mean you can add -6z and -2z. (-6z) + (-2z) = (-8z)

The final equation would be z^2-8z+12

I hope this helps!

Answer 4

z^2-8z+12
c=12 is >0 so both factors will have the same sign.
b=-8 is < zero so the factors will both be negative.

all yo uneed now are 2 numbers who's sum is -8 & who's product is 12

factors of 12 are
12*1 12+1=13
6*2 6+2=8
4*3 4+3=7

-6 & -2 are the right factors so
z^2-8z+12
(z-6)(z-2)=0
z-6=0
z=6

z-2=0
z=2

Answer 5

Set the equation to zero:
z^2-8z+12=0

Factoring:
(z-2)(z-6)=0
z-2=0
z-6=0
z=2 and 6

Check:
(2)^2-8(2)+12=0
4-16+12=0
-12+12=0
0=0

(6)^2-8(6)+12=0
36-48+12=0
-12+12=0
0=0

Answer 6

You need to use the quadratic equation. It should be in your mathbook. The quadratic equation finds the roots of a second order equation. Your equation must be in the form of

ax^2 + bx +c (which yours is)

Your a = 1
Your b = -8
Your c = 12

Then you use the formula -b + or - the sq root of b^-4ac all divided by 2a.

For you equation, the roots are z=6 and z=2
Thus, you end up with (z-6)*(z-2) as a factored equation.

Good luck!

Answer 7

z^2-8z+12 = (z-6)(z-2) '
it is always good to use the quadratic formula:

z= ( 8 +- sqrt( 64-4(12)) )/2
z= 6 or z = 2

Answer 8

If that's only factoring, then i assume this could be: x^2y (2x^2y - 3x^y - 20) when you consider that x^4y = x^2y * x^2y x^3y = x^2y * x^y. whilst multiplying exponents, you upload the exponents of words with a similar bases.