How do you find out the factors in the factoring complete in these problems?

Question

1. 3x3 – 27x2 + 60x
2. x2 – 7xy + 7ax – 49ay
3. x2 – 4x + 3
4. 8x2 – 2y2
This are practice problems that I am having problems with.

Answer 1

1. First factor out a 3x, so you have 3x(x^2-9x+20). To factor the rest, look for factors of the form (x+a)(x+b) where ab=20 and a+b=-9. The two numbers should be negative because their sum is negative and product is positive. You end up with a=-4 and b=-5, so the factorization is 3x(x-4)(x-5).

2. Factor an x out of the first two terms: x(x-7y). Then factor 7a out of the second two terms: 7a(x-7y). Note that what's left over is the same in each case. So your factorization is (x+7a)(x-7y).

3. Proceed as in the second step of the first problem: seek a factorization of the form (x+a)(x+b) where ab=3 and a+b=-4. You get (x-1)(x-3).

4. Factor out a 2: 2(4x^2-y^2). What's left is a difference of two squares. Use (a^2-b^2) = (a+b)(a-b) to obtain 2(2x+y)(2x-y).

Answer 2

1) 3x^3 - 27x^2 + 60x

Factor out 3x from each term, giving

3x(x^2 - 9x + 20)

You know the trinomial, if it can be factored, would become (x + a)(x + b) where the product ab = 20 and their sum = -9

The factors of 20 are 1,20,2,10,4,5 but only the combination of 4 & 5 will give the x term, so we have

3x(x - 4)(x - 5)


2) x^2 - 7xy + 7ax - 49ay

You have to keep your eyes open and look for patterns. Note that you can group the first two terms together and factor out an x giving x(x - 7y) and from the last two terms you can factor out 7a giving 7a(x - 7y) so the problem can be written as

x(x - 7y) + 7a(x - 7y)

Now we have only two terms in the expression and we can factor out a (x - 7y) from each of them

(x - 7y)(x + 7a)


3) x^2 - 4x + 3

This trinomial is easier than the trinomial in problem 1 in that the only factors of the constant term are 1, 3

(x - 1)(x - 3)


4) 8x^2 - 2y^2

Always look for a factor common to all terms in the expression, here you have the factor 2

2(4x^2 - y^2)

This binomial can be referred to as the difference of two squares: 4x^2 = (2x)(2x) and y^2 = (y)(y)

Look at a simpler example

x^2 - 1

and write it with an x term

x^2 + 0x - 1

The factored form should look like

(x + a)(x + b)

But, since the constant term = 1, both a and b must have the absolute value of 1, and in order to get o (zero) for the coefficient of the x term, one must be positive and the other negative.

(x + 1)(x - 1)

The difference of two squares is easy to factor

2(4x^2 - y^2) = 2(2x + y)(2x - y)

Answer 3

For the first one, you factor out a 3x, then factor as usual

For the 2nd, you factor an x out of (x2-7xy) and 7a out of (7ax-49ay).Then you get (x+7a)(x-7y)

For the 3rd, you get (x-1)(x-3)

For the 4th, factor out a 2, then factor to 2(2x+y)(2x-y)

Answer 4

1. 3x(x2-9x-20)
2 x(x-7y)+7a(x-7y)
3 X(x-4)+3
4 2(4x2-y2)