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

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embizze | High School Teacher | (Level 1) Educator Emeritus

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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:
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Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

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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.




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mchandrea | Student, College Junior | (Level 1) Salutatorian

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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.

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rachellopez | Student, Grade 12 | (Level 1) Valedictorian

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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.

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