Algebra II Factoring Question: y^6 + 7y^3 - 8?

Question

I can't seem to figure out the following Algebra II question:

Factor the polynomial:

y^6 + 7y^3 - 8

I have the answer, in the back of the book, but I can't seem to figure out the steps to get to the answer. Please help.

Thanks,

Bob

Answer 1

(y^6) + 7(y^3) - 8 = 0

let y^3 = x

Therefore,

(x^2) + 7x - 8 = 0

(x + 8) (x - 1) = 0

x = -8, x = 1

Therefore,

y^3 = -8, y^3 = 1



Which gives y = -2, y = 1

Answer 2

(y³ + 8) (y³ - 1)

Hint: let x = y³ and factor x² + 7x - 8

Answer 3

Write 2 sets of parentheses.

(_________)(_________) =

Look at the first term's factors.

y^6 = y^0 * y^6 = y * y^5 = y^2 * y^4 = y^3 * y^3 = y^4 * y^2 = y^5 * y = y^6 * y^0

Since the middle term is y^3, we know that we want y^3 * y^3 as the factors.  Put y^3 at the beginning of each set of parentheses.

(y^3_____)(y^3_____) =

Look at the last term's factors.  You want the pair that will add up to the middle coefficient: 7.

Multiples

1*-8 ==> 1 + -8 = -7 <== wrong pair

2*-4 ==> 2 + -4 = -2 <== wrong pair

4*-2 ==> 4 + -2 = 2 <== wrong pair

8*-1 ==> 8 + -1 = 7 <== right pair

Write 8 and -1 in the remaining blanks.

(y^3 + 8)(y^3 - 1) =

Notice that y^3 + 8 is the sum of 2 squares.

y^3 + 8 = (y)^3 + (2)^3

We know that a^3 + b^3 = (a + b)(a^2 - ab + b^2)

Given: y^3 + 8 = (y)^3 + (2)^3

Means: a = y, b = 2

Replace y^3 + 8 with the factored form.

(y^3 + 8)(y^3 - 1) =

(y + 2)(y^2 - 2y + 4)(y^3 - 1) =

Notice that y^3 - 1 is the difference of 2 squares.

y^3 - 1 = (y)^3 - (1)^3

We know that a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Given: y^3 - 1 = (y)^3 - (1)^3

Means: a = y, b = 1

Replace y^3 - 1 with the factored form.

(y + 2)(y^2 - 2y + 4)(y^3 - 1) =

(y + 2)(y^2 - 2y + 4)(y - 1)(y^2 + y + 1)

ANSWER: (y + 2)(y^2 - 2y + 4)(y - 1)(y^2 + y + 1)

Answer 4

=(y^3+8)(y^3-1)

Maybe this will help.

x=y^3

x^2+7x-8=

(x+7)(x-8)=

(y^3+8)(y^3-1)

Answer 5

think of this as a quadratic and rewrite it...

(y³)² + 7(y³) - 8

this is in the form x² + 7x - 8 where x is y³

so...

(y³ + 8)(y³ - 1)

Answer 6

Let's say that y^3=x

so:

x^2+7x-8 factors to:

(x-1)(x+8)

Plug in y^3 for x

(y^3-1)(y^3+8)

Those are special cases, a sum and difference of cubes.

Difference of cubes is:

(a^3-b^3)=(a-b)(a^2+ab+b^2)

Sum of Cubes is:

(a^3+b^3)=(a+b)(a^2-ab+b^2)

So:

(y^3-1)(y^3+8)=

(y-1)(y^2+y+1)(y+2)(y^2-2y+4)