# Can someone please show how to answer a couple of these questions? Thanks!

### 4 Answers | Add Yours

We are asked to fully factor the given polynomial:

A good rule of thumb is to factor out the greatest common factor of all of the terms. Then see if the remaining terms are in a recognizable form; e.g. the difference of two squares, the sum or difference of two cubes, a perfect square trinomial, a quadratic form, etc... A polynomial is fully factored in the reals if it is expressed as the product of linear or irreducible quadratic factors.

(1) `24x^3+18x^2 ` First factor out the greatest common factor `6x^2 `

` 6x^2(4x+3) ` which is fully factored.

(2) `2y^3-10y^2-12y ` First factor out the gcf `2y `

`2y(y^2-5y-6) ` Now factor the trinomial

`=2y(y^2-6y+1y-6) `

`=2y(y(y-6)+1(y-6)) `

`=2y(y-6)(y+1) ` which is fully factored.

(3) `-7m^3+28m^2-21m ` First factor out the gcf -7m:

`-7m(m^2-4m+3) ` Now factor the trinomial

` =-7m(m-3)(m-1)` which is fully factored.

(4) `4t^3-144t ` Factor out the gcf 4t

`4t(t^2-36) ` Recognize the binomial as the difference of two squares

`=4t(t+6)(t-6) ` which is fully factored.

(5) `c^4+c^3-12c-12 ` The gcf is 1, so try factoring by pairs:

`=c^3(c+1)-12(c+1) ` Now use the distributive property

`=(c+1)(c^3-12) ` Now we can consider the binomial to be the difference of two cubes (Some texts will stop at this point)

`=(c+1)(c-root(3)(12))(c^2+c root(3)(12) +root(3)(12^2)) `

(6) ` 6b^4+5b^3-24b-20 ` the gcf is 1 so factor by pairs

`=b^3(6b+5)-4(6b+5) `

`=(6b+5)(b^3-4) ` Again, in a year 1 Algebra course you could stop here or

`=(6b+5)(b-root(3)(4))(b^2+b root(3)(4)+root(3)(4^2)) `

(7) `a^3+6a^2-4a-24 ` The gcf is 1 so factor by grouping

`=a^2(a+6)-4(a+6) ` Use the distributive property

`=(a+6)(a^2-4) ` Recognize the binomial as the difference of two squares

`=(a+6)(a+2)(a-2) `

(8) `3m^3-15m^2-6m+30 ` Factor out the gcf 3

`3(m^3-5m^2-2m+10) ` Factor the polynomial using grouping

`=3(m^2(m-5)-2(m-5)) ` Use the distributive property

` =3(m-5)(m^2-2) ` In a first year course stop here or

`=3(m-5)(m+sqrt(2))(m-sqrt(2)) `

**Sources:**

To factor expressions, the easiest first step is to look for the greatest common factor of all the terms. Divide by that factor to get a simplified expression. For example in (35), the greatest common factor is 3.

3m^3 - 15m^2 - 6m + 30

Then see if you can factor anything else.

Never divide a whole expression by a variable to cancel it out because you will lose a solution.

You are asked to factor out the algebraic equations. To do this, you must always find first a common factor in each expression.

Take number 25 for example:

`-7m^3 + 28m -21m`

`7m (-m^2 + 4m - 3)`

Get the common factor which is 7m and put it on one side; and find factors for the equation in the parentheses

`7m (-m + 3)(m - 3)` This would be the answer. A completely factored form.

For number 27 (and the rest of the numbers, you could do the same) Here's another example.

`4t^3 - 144t`

`4t (t^2 - 36)` Get the common factor; and factor out the remaining equation.

`4t (t - 6) (t+6)` This is now the final factored form.

When you are factoring, you will look for the largest thing that each term has in common.

For number 21, the thing that both terms have in common is 6x^2. ` `` ` If you take this out of each term you are left with **6x^2(4x+3)**. This cannot be factored any further so that is your answer.` `

The same is with number 23, even if it has more terms. The thing that all three have in common is 2y. When you factor you are left with 2y(y^2 - 5y - 6). However you are not done because the second term can still be factored one more time. Y times y will give you y^2, so you know both factored terms must include y. To get a -6 at the end but a 5y in the middle you must include a -6 and +1 in your factors. The two new factors you will get will be (y - 6)(y + 1) which you can redistribute to check your work. The final answer is **2y(y - 6)(y +1)**.

I hope that helps but if you have any questions I will gladly answer them.