Factoring help?

[TheAmandalarson]

How do I factor 4x^3+21x^2+21x+4 if (x+4) is one of its factors?
I started off by doing long division and I got 4x^2+5x+1

 

[Deleted member 4993]

How do I factor 4x^3+21x^2+21x+4 if (x+4) is one of its factors?
I started off by doing long division and I got 4x^2+5x+1

Do you know how to factorize "quadratic function" - like ax[sup:l8pzku97]2[/sup:l8pzku97] + bx +c ?

 

[TheAmandalarson]

How do I factor 4x^3+21x^2+21x+4 if (x+4) is one of its factors?
I started off by doing long division and I got 4x^2+5x+1

Do you know how to factorize "quadratic function" - like ax[sup:7li5vhrp]2[/sup:7li5vhrp] + bx +c ?

Yes, x=-b+-square root of b^2-4ac over 2a, right?

 

[Denis]

Yes, x=-b+-square root of b^2-4ac over 2a, right?

Or simply factor: (4x + 1)(x + 1) = 0

 

[mmm4444bot]

I started off by doing long division and I got 4x^2 + 5x + 1

This is the correct quotient; it is factorable.

 

[mmm4444bot]

Do you know how to factorize "quadratic function"

Yes, x=-b+-square root of b^2-4ac over 2a, right?

Not quite.

The Quadratic Formula finds the roots of a 2nd-degree polynomial. That process is not "factorizing".

However, once you know the two roots, you can then write a factorization as (x - root1)(x - root2).

There is another method, called "factoring by grouping", that I like.

You first multiply A times C. You then look for two numbers whose product is AC and whose sum is B.

AC = 4

B = 5

Two numbers whose product is 4 and whose sum is 5 are 1 and 4

Rewrite the middle term of the polynomial, using these two numbers.

4x^2 + 4x + x + 1

Group the first two terms, group the last two terms, and factor.

(4x^2 + 4x) + (x + 1)

4x(x + 1) + (x + 1)

The factor (x + 1) appears on both sides of the red plus sign; therefore, it can be factored out.

(x + 1)(4x + 1)

We have the following result:

4x^3 + 21x^2 + 21x + 4 = (x + 4)(x + 1)(4x + 1)

Cheers ~ Mark