How do you solve a problem like these?

Question

1. 2x^2 - 18

2. 3x^2 -27x + 60

3. 12x^2 + 8x -15

Could you please explain how you did them?

Answer 1

Way too much explanation...

First one is 2x^2 - 9 take out the 2

2(x^2 - 9) and then factor the parentheses

2(x+3)(x-3)

Second is 3x^2 - 27x + 60  Take out the 3

3(x^2 - 9 + 20) then factor the parentheses

3(x - 5)(x - 4)

Last one is unfactorable.  Try the above method in number two to do the last one and you'll find you cannot factor it out no matter what you do.

Answer 2

2x² - 18

First factor out 2

2(x² - 9)

Next find out if what is inside the parentheses can be factored.

(x² - 9) is the same as (x² + 0x - 9)

So we need to identify two numbers (a and b) whose sum is 0 and whose product is -9.

To solve for these numbers, we set up two equations:

a + b = 0

ab = -9 so a = -9/b

Substitute "-9/b" for "a" in the first equation to get

-9/b + b = 0

(-9 + b²)/b = 0

multiply both sides by b

-9 + b² = 0

b² = 9

b = +3 and -3

So, (x² - 9) factors out to

(x+3)(x-3)

Finally, our complete equation now becomes

2(x+3)(x-3)

This is as far as we can factor out.

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I will get you started with problem 2,

3x² - 27x + 60

Factor out 3 from all three terms

3(x² - 9x + 20)

Next see if you can factor out what is inside the parentheses.

(x + ?)(x + ?)

The unknown numbers have a sum of -9 and a product of +20. You should be able to figure out this one.

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12x² + 8x - 15

This one is more complicated.

Factor out 12.

12[x² + (8/12)x - 15/12]

(x + ?)(x + ?)

We need two numbers whose sum is +8/12 and whose product is -15/12. To make this easier, since both numbers are divided by 12, we can temporarily ignore the 12. Therefore, we need two numbers whose sum is +8 and whose product is -15.  I believe that when you analyze this, it is not factorable.