Ok guys, sorry if im posting this in the wrong section; dont know where quadratics belong.

Anyway, so I have a maths test next week, and my maths teacher handed out practice tests (from years back) for us to get a look at what we're dealing with. Anyway, im going pretty well with it, except theres 1 question im stuck at. Could you please help me do it?

Question: Factorize each of the following expressions, over R if necessary.

1) $16x^2-81$
2) $x^2+4x-5$
3) $-7(x-3)^2+63$
4) $2x^2-22x+8$
5) $7(9+5)^2 -6(9+5)-1$

Thanks

2. Originally Posted by mcaelli
Ok guys, sorry if im posting this in the wrong section; dont know where quadratics belong.

Anyway, so I have a maths test next week, and my maths teacher handed out practice tests (from years back) for us to get a look at what we're dealing with. Anyway, im going pretty well with it, except theres 1 question im stuck at. Could you please help me do it?

Question: Factorize each of the following expressions, over R if necessary.

1) $16x^2-81$
2) $x^2+4x-5$
3) $-7(x-3)^2+63$
4) $2x^2-22x+8$
5) $7(9+5)^2 -6(9+5)-1$

Thanks
Hi mcaelli,
1) $16x^2-81$

This is the classic difference of two squares. Use the following model to factor:

$a^2-b^2=(a-b)(a+b)$

2) $x^2+4x-5$

For this one, think of two numbers whose product is -5 and whose sum is 4. You'll discover that there is only one solution to that question: 5 and -1

$(x+5)(x-1)$

3) $-7(x-3)^2+63$

Try expanding everything and collecting terms to get a neat quadratic expression. Then, we'll try to factor it. Or, better still, do what stapel suggests in her post. It's cleaner.

4) $2x^2-22x+8$

Observe that the expression has a common monomial factor.

$2(x^2-11x+4)$

5) $7(9+5)^2 -6(9+5)-1$

This is a strange little booger!

Let x = (9+5), then $7x^2-6x-1$

$(7x+1)(x-1)$

Now back substitute to get $[7(9+5)+1)][(9+5)-1]$

I really don't know what you might want to do next since everything here is numbers.

3. Originally Posted by mcaelli
Factorize each of the following expressions....

1) $16x^2-81$
To learn how to factor a difference of squares, try here.

Originally Posted by mcaelli
2) $x^2+4x-5$
To learn how to factor regular quadratics, try here.

Originally Posted by mcaelli
3) $-7(x-3)^2+63$
To learn how to take the common factor out front, try here. This will leave you with:

. . . . . $-7\left((x\, -\, 3)^2\, -\, 9\right)$[/quote]

Then factor the remaining difference of squares.

Originally Posted by mcaelli
4) $2x^2-22x+8$
Take the common factor of "2" out front. What remains does not factor "over the integers". However, you are supposed to plow on regardless. So work backwards from the zeroes:

Set the quadratic equal to zero, plug it into the Quadratic Formula, and find the roots, being some (messy) numbers "a" and "b".

Then note that "x^2 - 11x + 4 = 0 for x = a, x = b" means that "(x - a)(x - b) = 0", so the factors must be (x - a) and (x - b).

4. Sweet guys, thanks alot. Still a bit unsure about Q5 but got 1-4 done so im pretty happy with it .

5. Originally Posted by mcaelli
Sweet guys, thanks alot. Still a bit unsure about Q5 but got 1-4 done so im pretty happy with it .
Yeah, #5 was a bit strange, I agree. But I factored it as requested. Notice the substitution I made to make it easier. This one is just arithmetic, but the instructions were to factor. That's why I left it in that form.

Good Luck. Glad we could help.