Question

ok, im not good in math but trying! I am factoring the given polynomial 8x2 -6x -9 I know you AC and get 72 (here is my problem) what is [] . [] = 72 and []+[] = -6 (where do you start to even guess what two numbers will do this????

Answer 1

What are the factors of 72?

Answer 2

72=2*36=2*2*18=2*2*2*9=2*2*2*3*3
That gives you the following numbers to play with:
2,3,4,6,8,9,12,18,24,36

Answer 3

yes, what product is 72 as well as is the sum of -6?

Answer 4

We also know that to get an 8x^2 term we need either (x+something)(8x+somethingelse) or (2x+something)(4x+somethingelse)

Answer 5

sorry, yes, I didnt know how to make the exponent (8x^2)

Answer 6

Let's write that as \[(2x+a)(4x+b)=8x^2-6x-9\]
But\[(2x+a)(4x+b)=8x^2+4ax+2bx+ab = 8x^2 + (4a+2b)x+ab\] and we need (4a+2b)=-6 and ab = -9 so let's try 3 and -3 (their product being 9). 4(3)+2(-3)=6 so try 4(-3)+2(3) = -6 so our factored polynomial is \[ (2x-3)(4x+3)\] (a little thought will quickly eliminate -1,9 or 1,-9 as possibilities for a and b)

Answer 7

So the key was not what product is 72 as well as sum of -6, but product of -9 and then weighted sum of -6 when multiplied by 2 and 4.

Answer 8

YOU ROCK!!!!! I guess I was trying to use the wrong formula for this.... this is going to be the longest 16 weeks of my life!!!!

Answer 9

I have to confess that I've always found factoring polynomials to be very tedious :-)
You were headed in the right direction if you didn't have that pesky 8 in front of x^2. Let's say the problem was really \[x^2-6x +9\]Then you would have to find ab such that a+b=-6 and ab = 9. You could solve that or just spot that -3 + -3 = -6 and -3 * -3 = 9 like you were trying to do with the original problem and the factors would be \[(x-3)(x-3)=x^2-6x+9\]

Answer 10

yeah, Im trying to take this algebra class online and kirk trigstead is the books author; his videos do not help me all that great... so of course, I am trying my hand at youtube to find some more videos to understand it!

Answer 11

You also could divide the original polynomial by 8 to put it in standard form, then complete the square, and finally multiply by 8 once again:
\[x^2-\frac{6}{8}x-\frac{9}{8}=0\]\[x^2-\frac{6}{8}x=\frac{9}{8}\]Take half of the x coefficient, square it, add it to both sides\[x^2-\frac{6}{8}x+\frac{3^2}{8^2}=\frac{3^2}{8^2}+\frac{9}{8}\]Now write the polynomial as a square\[(x-\frac{3}{8})^2= \frac{81}{64}\] then take the square root of each side\[(x-\frac{3}{8})=\pm\frac{9}{8}\] then solve for x. x = 3/2, x = -3/4. Now let's put those back together in our original equation.

Our "simplified" equation was \[x^2-\frac{6}{8}x-\frac{9}{8}=0\] which factored to\[(x-\frac{3}{2})(x+\frac{3}{4}) = 0\] If we multiply that by 4*2=8 (remember, we divided by 8), we have \[2*(x-\frac{3}{2})*4*(x+\frac{3}{4}) = 0\]and that in turn simplifies to\[(2x-3)(4x+3)=0\] and our factored expression would just be the left hand side of that. Because we set it equal to 0 before completing the square, and have maintained that equality, we can now remove the = 0 without fear that we've bollixed something up.

I'm not suggesting that one would necessarily want to do the factoring by completing the square, but I often find it helpful to my understanding to see how the different methods might work.