I'm having trouble working out these problems. Could someone please assist?

with this first one do I factor the x^2 + 3xy - 18y^2 first?

x - 3y * x^2 + xy - 30y^2 /
x^2 + 3xy - 18y^2

3a^3b^2 * 4a^2 + 2ab + b^2
8a^3 - b^3 12ab^4

3y + 12 / 9y+36
8y^3 16y^3

Susan I see these are rational fractions. I'm not sure though, do you mean for them to be poly's like p(x)/q(x)?
If they are, then factor them to find common terms. Are you sure the numerator of the first one is written correctly? You can factor in either order, but keep track of numerator and denominator factors. Cancel common ones.

Don Don answered
8 years ago

That craz question

Explain Bot Explain Bot answered
8 months ago

Yes, for the first problem, you can factor the numerator and denominator separately. Let's start with the numerator:

x - 3y * (x^2 + xy - 30y^2)

To factor the quadratic expression x^2 + xy - 30y^2, you can look for two numbers that multiply to -30 and add up to the coefficient of the middle term, which is 1 in this case. In this case, the numbers are 6 and -5 because 6 * (-5) = -30 and 6 + (-5) = 1.

So the expression x^2 + xy - 30y^2 can be factored as (x + 6y)(x - 5y).

Putting it all together, the numerator becomes (x - 3y)(x + 6y)(x - 5y).

For the denominator, x^2 + 3xy - 18y^2, you can use the same factoring process. Look for two numbers that multiply to -18 and add up to the coefficient of the middle term, which is 3 in this case. The numbers are 6 and -3.

So the expression x^2 + 3xy - 18y^2 can be factored as (x + 6y)(x - 3y).

Therefore, the expression x^2 + 3xy - 18y^2 can be simplified as (x - 3y)(x + 6y)(x - 5y) / (x + 6y)(x - 3y).

Notice that the (x - 3y) and (x + 6y) terms cancel out in the numerator and denominator, so the final simplified expression is (x - 5y).