factoring- Help?

Question

Help with these factoring problems:

1. x^2 + 6x + 9y^2

2. x^3 - 25x

3. 18m^2 - 8y^2

4. 2a^5 - 162a

Answer 1

1) x^2 + 6x + 9y^2

This isn't easy to see, but you should first complete the square with the x terms.

x^2 + 6x + 9 + 9y^2 - 9

Now, factor.

(x + 3)^2 + 9(y^2 - 1)

Now, factor the second part as a difference of squares.

(x + 3)^2 + 9(y - 1)(y + 1)

2) x^3 - 25x

First things first, factor an x out.

x(x^2 - 25)

Now, factor as a difference of squares.

x(x - 5)(x + 5)

3) 18m^2 - 8y^2

First, factor out a 2.

2(9m^2 - 4y^2)

Now, factor as a difference of squares. Note that 9 and 4 are square numbers, and m and y are both squared (and therefore square numbers), so this becomes

2(3m - 2y) (3m + 2y)

4) 2a^5 - 162a

Factor out 2a.

2a(a^4 - 81)

Now, factor as a difference of squares. Note a^4 is a perfect square because a^2 times a^2 is a^4.

2a(a^2 - 9)(a^2 + 9)

Now, factor the a^2 - 9 as a difference of squares.

2a(a - 3)(a + 3)(a^2 + 9)

Answer 2

Let's see,

a. x^2 + 6xy + 9y^2 = (x+3y)(x+3y) Assuming the middle term is 6xy

b. x^3 - 25x = x(x^2 - 25) = x(x+5)(x-5)

c. 18m^2 - 8 y^2 = 2((3m)^2 - (2y)^2))
= 2[(3m+2y)(3m-2y)]

d. 2a^5 - 162a = 2a(a^4-81)= 2a(a^4-(3)^4)

2a[(a^2-9)(a^2+9)]

Good luck!

Answer 3

1. I believe your middle term should be 6xy
(x + 3y)(x+3y)

2. x(x^2-25)
x(x+5)(x-5)

3. 2(9m^2-4y^2)
2(3m+2y)(3m-2y)
4. 2a(a^4-81)
2a(a^2+9)(a^2-9)
2a(a^2+9)(a+3)(a-3)

Answer 4

1. (3y) (3y) + x(x + 6)

2. x( x^2 - 25)

3. 2(9m^2 - 4y^2)

4. 2a(a^4 - 81) = 2a(a^2 + 9) (a^2 - 9)

Answer 5

1.
2. x(x-5)(x+5)
3. 2(9m^2-4y^2)
4. a(2a^4-162)

Answer 6

(x+3y)^2

x(x^2-25)= x(x-5)(x+5)

2(9m^2-4y^2)= 2(3m+2y)(3m-2y)

2a(a^4-81)= 2a(a^2-9)^2=2a(a-3)(a+3)

Answer 7

1. (x+3y)(x+3y)

2. x(x+5)(x-5)

3. 2(3m+2y)(3m-2y)

4. 2a(a+3)(a+3)(a+3)(a-3)