Factor completely 21x3 + 35x2 + 9x + 15. (3x - 5) (7x2 - 3) (3x - 5) (7x2 + 3) (3x + 5) (7x2 - 3) (3x + 5) (7x2 + 3)

Cubic = (3x + 5) (7x^2 + 3)

Step-by-step explanation:

If you make up two groups of 2, you might be able to factor this using the distributive property. Let's try it.

First group: 21x^3 + 35x^2 Take out 7x^2 as a common fact

First group: 7x^2 (3x + 5)

Second group: 9x + 15 The HCF is 3

Second group: 3 (3x + 5)

Now put your factors together in one long string.

Cubic = group 1 + group 2

Cubic = 7x^2 (3x + 5) + 3 (3x + 5)

Note: If you had something like 7x^2 * y + 3 * y then you should see that the factors are y * (7x^2 + 3). So to solve the cubic, you should observe that (3x + 5) is a common factor.

Let y = 3x + 5

y (7x^2 + 3) Now substitute back for the y.

Cubic = (3x + 5) (7x^2 + 3)

(7x^2+3) (3x+5)

Step-by-step explanation:

21x^3 + 35x^2 + 9x + 15.

We will do factor by grouping

Rearranging the terms so I have the terms with a factor of 3 first and 5 last

Factor a 3x from the first two terms and 5 from the last two terms

21x^3 + 9x + 35x^2 + 15

3x (7x^2 + 3) + 5 (7x^2+3)

Now factor out a (7x^2+3)

(7x^2+3) (3x+5)