Factoring Help!

**badandy328** · October 17th 2007, 03:40 PM

I'm having trouble doing these problems:

1. Solve the equation

What are the 2 solutions? I get -1 and 4 but apparently I'm wrong

2. The polynomial

can be factored into the product of three polynomials, A*B*C. Find A B and C.

No clue how to do this one.

3. The polynomial

can be factored into the product of two polynomials, A*B. Find A and B.

Now I got one of the factors to be 9x-8y but i can't find the other.

**Krizalid** · October 17th 2007, 03:44 PM

Originally Posted by badandy328

I'm having trouble doing these problems:

1. Solve the equation

What are the 2 solutions? I get -1 and 4 but apparently I'm wrong

2. The polynomial

can be factored into the product of three polynomials, A*B*C. Find A B and C.

No clue how to do this one.

3. The polynomial

can be factored into the product of two polynomials, A*B. Find A and B.

Now I got one of the factors to be 9x-8y but i can't find the other.

For the first one, multiply both sides by $6(x+1)(x+2)$

For the third one, use the following identity $a^3-b^3=(a-b)(a^2+ab+b^2)$

I'm too lazy now for the second one

**badandy328** · October 17th 2007, 03:50 PM

Originally Posted by Krizalid

For the first one, multiply both sides by $6(x+1)(x+2)$

For the third one, use the following identity $a^3-b^3=(a-b)(a^2+ab+b^2)$

I'm too lazy now for the second one

For the first one though I end up with 0= x^2+3x-16
I need to get two solutions, and I don't think this could be factored. Am I doing something wrong?

**Soroban** · October 17th 2007, 04:12 PM

Hello, badandy328!

2. The polynomial . $81x^3 + 486x^2 - x - 6$ can be factored
into the product of three polynomials, $A\cdot B\cdot C$
Find $A,\,B\text{ and }C.$

Factor "by grouping"

$\text{We have: }\;81x^2\underbrace{(x+6)}_{\text{common}} - \underbrace{(x + 6)}_{\text{factor}}$

$\text{Factor: }\;(x + 6)\underbrace{(81x^2 - 1)}_{\text{diff. of squares}}$

$\text{Factor: }\;(x + 6)(9x-1)(9x+1)$

**badandy328** · October 17th 2007, 04:19 PM

thanks! All i need now is number 1. I can't figure it out

**Soroban** · October 17th 2007, 04:23 PM

Hello, badandy328!

Did you take Krizalid's advice for the first one?

1. Solve the equation: . $\frac{1}{x+1} - \frac{1}{x+2} \;=\;\frac{1}{6}$

Multiply through by $6(x+1)(x+2)$

. . $6(x+1)(x+2)\cdot\frac{1}{x+1} - 6(x+1)(x+2)\cdot\frac{1}{x+2} \;=\;6(x+1)(x+2)\cdot\frac{1}{6}$

Reduce: . $6(x+2) - 6(x+1) \;=\;(x+1)(x+2)$

Expand: . $6x + 12 - 6x - 6 \;=\;x^2 + 3x + 2$

We have the quadratic: . $x^2 + 3x - 4 \;=\;0$

. . which factors: . $(x - 1)(x + 4) \;=\;0$

. . and has roots: . $x \:=\:1,\,-4$

**badandy328** · October 17th 2007, 04:28 PM

Originally Posted by Soroban

Hello, badandy328!

Did you take Krizalid's advice for the first one?

Multiply through by $6(x+1)(x+2)$

. . $6(x+1)(x+2)\cdot\frac{1}{x+1} - 6(x+1)(x+2)\cdot\frac{1}{x+2} \;=\;6(x+1)(x+2)\cdot\frac{1}{6}$

Reduce: . $6(x+2) - 6(x+1) \;=\;(x+1)(x+2)$

Expand: . $6x + 12 - 6x - 6 \;=\;x^2 + 3x + 2$

We have the quadratic: . $x^2 + 3x - 4 \;=\;0$

. . which factors: . $(x - 1)(x + 4) \;=\;0$

. . and has roots: . $x \:=\:1,\,-4$

OHHH i forgot to bring over that subtraction sign. I got -1 and 4. Thank you so much.

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