# `3x^2 + 17x + 10` Factor the trinomial. If the trinomial cannot be factored, say so.

Asked on by nick-teal

### Textbook Question

Chapter 5, 5.2 - Problem 36 - McDougal Littell Algebra 2 (1st Edition, Ron Larson).
See all solutions for this textbook.

Educator Approved

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

`3x^2+17x+10`

aka

`ax^2 + bx + c  `

because there is a number in front of a we multiply it with c

so the first step is:

3 x 10 = 30

Then we find factors of 30 that equal b (17)

Factors of 30 are 1 x 30, 2 x 15, 3 x 10, 5 x 6

As we can see 2 and 15 are the ones that could add up to 17, so we replace b with these:

`3x^2+15x+2x+10 `

Group:

`(3x^2+15x)+(2x+10)`

Factor out common factors:

`3x(x+5)+2(x+2)`

`(3x+2)(x+5)`

Now set these parentheses equal to 0 and solve

3x + 2 =0

3x = -2

`x = -2/3`

x + 5=0

`x = -5`

Educator Approved

udonbutterfly | Student, College Freshman | (Level 1) Valedictorian

Posted on

Okay Lets start with multiply 3 by 10 which will equal...

10X3=30

Now we have to have multiples of 30 that when add/subtracted will gives us the middle term so 17x

2x15=30  15+2 this will gives us 17. Now replace these two numbers with the middle term

3x^2+17x+10

(3x^2+2x)+(15x+10) Now take out the number that each expression is divisible by

x(3x+2)+5(3x+2) The expressions in the parentheses should match because it will only be one expression for the final equation. The second expression will be the combination of the two terms you got when you took out the number the expressions were divisible by

(3x+2)(x+5) And you got the answer!

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