Need help with Factors of Polynomials ASAP. Thanks a bunch!

Question

1. Find the two errors in the work shown below. Explain the errors and then enter the correct solutions, showing the correct steps.

1. 6x³ - 7x² - 3x = 0
2. x(6x² - 7x - 3) = 0
3. x(6x² - 9x + 2x - 3) = 0
4. x[3x(2x - 3) + 1(2x - 3)] = 0
5. x(3x + 1)(2x - 3) = 0
6. 3x + 1 = 0, 2x - 3 = 0
7. 3x = -1, 2x = 3
8. x = -1/3, 2/3

2. Solve by factoring 6x³ - 16x² - 32x = 0. Enter your work.

3. Solve by factoring 9x² - 25 = 0. Enter your work.

Answer 1

Hello, Kenneth,

1. The point where we see: x(3x + 1)(2x - 3) = 0 on line 5 looks good, It is telling us that any of the three factors can equal zero. But the x = 0 is missing on line 6. It is a possible answer, so should be included. Same in line 8. It should read x = -1/3, 2/3, 0.

Those are the only errors I can can find.

Factoring 6x³ - 16x² - 32x = 0

2*(3x³ - 8x² - 16x) = 0

2x*(3x²  - 8x - 16) = 0

2x*(3x²  - 8x - 16) = 2x*(x-4)(3x+4) = 0

Solutions: 2x = 0, x = 0

x-4 = 0; x = 4

3x+4 = 0; x = -3/4

Factoring 9x² - 25 = 0

9x² - 25 = 0

(3x-5)(3x+5) = 0

x = 5/3, -5/3

Bob

Answer 2

1.

Between lines 5 and 6, they forgot that x = 0 is also an answer. (Three answers, not two.)

Between lines 7 and 8, they divided wrong. 2x = 3 would lead to x = 3/2 (not 2/3).

2.

Start by factoring out the GCD. Then you will be left with a quadratic polynomial which you can factor using "reverse FOIL".

3.

Factor using difference of two squares. 9x² = (3x)² and 25 = 5²