Question

One of the factors of 6x(Squared) + x - 12 =?
Then they explain it with this (which makes no sense to me):
Compare 6x
2 +x−12 with ax2 +bx+c. List the integer
pairs whose product is ac = −72.
−1, 72 1, −72
−2, 36 2, −36
−3, 24 3, −24
−4, 18 4, −18
−6, 12 6, −12
−8, 9 8, −9
Select the pair that adds to b = 1; that is, −8 and 9. Now,
6x
2 + x − 12 = 6x
2 + 9x − 8x − 12
= 3x(2x + 3) − 4(2x + 3)
= (3x − 4)(2x + 3).

Can someone please tell me how this makes sence? and is there a squared button on the computer?

Answer 1

take another simpler example

Answer 2

\(\large x^2 + 5x + 6\)

Answer 3

do u knw how to factor that ?

Answer 4

If you don't want to use the equation editor or \(\LaTeX\) to format the equations, you can indicate exponents by using ^ (which is usually found as the shifted value of the 6 key on a US keyboard, at least)

Compare
\[6x^2 +(1) x-12\]\[ax^2+bx + c\]

Isn't it clear that to make the two equivalent, we'll need to have \[6=a\]\[1=b\]\[-12=c\]?

Answer 5

Now, to factor this bad boy, we multiply \(a*c = 6*(-12) = -72\)
Next, we look for a pair of factors of \(-72\) such that the two factors when added together = 1, which is the coefficient of the middle term. As the problem says, \(-8,9\) is such a pair, because \(-8*9 = -72, \text{ and } -8+9=1\)

Answer 6

Next, we rewrite the equation using that information we just found. Instead of writing \(x\) in the middle, we'll write \(-8x +9x\) which is perfectly valid as \(-8x+9x=1x\), right?

\[6x^2-8x+9x-12\]Now we are going to put parentheses around each pair of terms:\[(6x^2-8x) + (9x-12)\]Again, no change to the value. Now factor the expression inside each set of parentheses:
\[2x(3x-4) + 3(3x-4)\]Any question about that?

Answer 7

Both of those terms have \((3x-4)\) as a common factor, so we can factor it out, which gives us:
\[(3x-4)(2x+3)\]

Now to check our work, we can multiply this out:
\[(3x-4)(2x+3) = 3x(2x+3)-4(2x+3)\]\[\qquad = 3x*2x+3x*3 - 4*2x -4*3\]\[\qquad=6x^2+9x-8x-12\]\[\qquad=6x^2+1x-12\]\[\qquad = 6x^2+x-12\]And that's what we started with, so our factoring must be correct.

Answer 8

ok thank you i am going to try to understand that

Answer 9

ok so why do you multiply ac to factor it? sorry i havn't taken algebra since high school :(

Answer 10

oh and thank you again...i am starting to understand the nature of this question!

Answer 11

The reason why you multiply a*c to find the factors in the middle:

Say we have two binomials we are going to multiply:
\[(mx+n)(px+q)\]\[\qquad =mx(px+q) + n(px+q) = mpx^2 + mqx + npx + nq\]\[\qquad = mpx^2 + (mq+np)x + nq\]
Comparing that with our polynomial we want to factor, \(ax^2+bx+c\)
\[a = mp\]\[b=mq+np\]\[c=nq\]
\[ac = mp*nq= mnpq\]

Notice that \(b=mq+np\) and both \(mq\) and \(np\) are factors of \(ac = mnpq\)

So, if we multiply \(a*c\) and pick a set of factors that add up to \(b\), we can arrange our equation into factored form.