[carchemishk]

25x^2+80x+64


15z^2+14z-8

write the expression in lowest term
2x+2
______
6x^2+16x+10

[Unknown008]

You need to know how to factorise, according to the numbers given.

1.
See +64 and +80x? That shows that you have to find positive factors of 64, which are +1, +2, +4, +8, +32 and +64.



By trial and error, you'll have;



using the +8 factors of +64, and the factors of 25.

2.Could you try that out now?

3. Factorise the denominator and the numerator, you'll see that something's will cancel out, giving a simpler fraction.

[Perito]

write the expression in lowest term

This appears to be about factoring. Let's take the first one.



1. Note that 25 and 64 are both perfect squares. 25 = 5*5; 64 = 8*8. 8 and 5 are, therefore, prime candidates for coefficients in the factored terms.

2, All terms are positive. Therefore, it will factor into something like this:

where a and c aren't necessarily different. In fact, since you've got a perfect square, a and c may be equal and b and d may be equal.

... And, I see that Uknown008 posted the answer. Make sure you understand what he did.

Let's look at the second problem.



1. 15 and 8 aren't perfect squares so we can't hope for anything simple like in #1. 15=3*5; 8=2*2*2. Since 15 has only two factors, we can expect 3 and/or 5 to be a coefficient (but only because problems are usually crafted this way). Also, 2 and 4 (2*2) are likely candidates for coefficients.

Since the signs of the terms are +, +, - we would expect to factor using this form:



of course because must be greater than ad because the second term is positive.

Now let's look at #3.



You can factor a 2 out of the numerator and the denominator. That would probably be a good idea right off the bat, so as to simplify things a tiny bit.



The 2 will divide out. Now it appears that we have an (x+1) in the numerator. Since the denominator is a quadratic expression, we should probably factor it to see if an (x+1) term appears that we can cancel out. Because the coefficients are all positive, we will again expect to factor the denominator into the form: and, in fact, we'll expect (if the problem was written the way most problems were written),

Give it a whirl.