1. factor the expression x^3 + 8y^3

2. factor x^9 - 27y^6

3. factor 4x^2-9

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- November 6th 2008, 08:01 AM #1

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- November 6th 2008, 09:20 AM #2

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## Solutions

1. factor the expression x^3 + 8y^3

2. factor x^9 - 27y^6

3. factor 4x^2-9

1.Well, since the first one is the sum of two cubes, you write it of the form:

(a + b)(a^2 - ab + b^2).

The cube root of x^3 is x because x*x*x = x^3.

The cube root of 8y^3 is found by taking the cube root of 8 and the cube root of y^3 and multiplying them together.

So, cube root of 8 is 2 and cube root of y^3 is y. 2y*2y*2y = 8y^3.

As per the above terms, we should find a^2, ab and b^2.

Allow a=x and b=2y

a^2 = x*x = x^2

b^2 = 2y*2y = 4y^2

ab = x*2y = 2xy

Now, we just use substitution:

(x + 2y)(x^2 - 2xy + 4y^2). and we are done.

- November 6th 2008, 10:30 AM #3

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## Solution 2

1. factor the expression x^3 + 8y^3

2. factor x^9 - 27y^6

3. factor 4x^2-9

1.Well, since the second one is the difference of two cubes, you write it of the form:

(a - b)(a^2 + ab + b^2).

The cube root of x^9 is x^3 because x^3*x^3*x^3 = x^9. You add the exponents when the terms are multiplied.

The cube root of 27y^6 is found by taking the cube root of 27 and the cube root of y^6 and multiplying them together.

So, cube root of 27 is 3 and cube root of y^6 is y^2. 3y^2*3y^2*3y^2 = 27y^6. The exponents are added, the base remains the same, and the coefficients are multiplied.

As per the above terms, we should find a^2, ab and b^2.

Allow a=x^3 and b=3y^2

a^2 = x^3*x^3 = x^6

b^2 = 3y^2*3y^2 = 9y^4

ab = x^3*3y^2 = 3(x^3)(y^2)

Now, we just use substitution:

(x^3 - 3y^2)(x^6 + 3(x^3)(y^2) + 9y^4). and we are done.

- November 6th 2008, 10:33 AM #4

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## Solution 3

3. factor 4x^2-9

This is the difference of squares, so it is of the form: (a + b)(a - b)

Take the square root of 4x^2 to find a and the square root of 9 to find b

square root of 4 is 2 and square root of x^2 is x

So, 2x is the square root of 4x^2 --> 2x*2x = 4x^2

square root of 9 is 3 --> 3*3 = 9

Let a = 2x and b = 3

Now just substitute:

(2x + 3)(2x - 3), and we are done