Factor this: 20n^2 + 34n + 6?

Question

n^2 means n squared.

Answer 1

We always look for a monomial factor first, and in this case there is one:

20n^2 + 34n + 6 = 2(10n^2 + 17n + 3)

Now we factor by grouping. 10*3 = 30. Is there a pair of numbers with 30 as a product but 17 as a sum? Yes, 15 & 2.

2(10n^2 + 17n + 3) = 2(10n^2 + 15n + 2n + 3)
= 2[ ( 10n^2 + 15n ) + ( 2n + 3 ) ]
= 2[ 5n( 2n + 3) + (2n + 3) ]
= 2[ (2n + 3)(5n + 1) ]
= 2(2n + 3)(5n + 1)

You can test the answer by multiplying the factors back together: when like terms are added, the result is the original polynomial.

Hope that helps!

EDITED TO ADD: JeffU's answer is *not* right. The factors he provided ( (n+.2)(n+1.5) ) do not recreate the original polynomial when multiplied together. (The two polynomials both have the same roots, but that's not the same thing.) The problem with using the quadratic formula in place of factoring is that many different quadratics can have the same roots, for instance:

x^2 - 4 = 0
2x^2 - 8 = 0
3x^2 - 12 = 0
4x^2 - 16 = 0

All four equations have the same roots (x = 2 or x = -2) but the quadratic expressions on the left side have different factored representations.

Answer 2

first you have to divide by 2

than you'll get 2 x (10n^2 + 17n + 3)

than the factors are 2 x (5n + 1) x (2n + 3)

to check if it is right 2 x ( 10n^2 + 15n + 2n + 3)
which is 2 x (10n^2 + 17n + 3) which is 20n^2 + 34n + 6

Answer 3

20n² + 34n + 6

To make it easy, factor out 2 first
= 2(10n² + 17n + 3)

Now it is easier to factor:
= 2(5n + 1)(2n + 3)

That is the complete factoring of 20n² + 34n + 6.
^_^

Answer 4

Using the Quadratic Formula, we find that n = -.2 and n = -1.5.

So the factors are (n+.2)(n+1.5)=0

........ Sorry, Many are pointing out that I made a mistake. I forgot to throw in 20 as my factor. My final answer is 20(n+.2)(n+1.5)= 20n^2 + 34n + 6. That should make you happier.

Answer 5

20n^2 + 34n + 6
2(10n^2 + 17n + 3)
2(2n + 3)(5n + 1)

Answer 6

Is Jeff U right?

In order to use the quadratic equation, should not the equation be set to equal zero? Since it is just a formula, I believe we will have to solve it like a monomial equation:

20n^2+34+6 -----we have to factor out the commonality
2 (10n^2+17n+3) -----now solve
2 (5n+1)(2n+3)

Is that it?

Answer 7

Each term has a common factor of 2.
2[10n^2 + 17n + 3]

To factor a trinomial (Ax^2 + Bx + C), multiply A times C, and find two factors of that product whose sum is equal to B.
10 * 3 = 30; factors of 30 that add to 17 are 15 and 2.
Substitute 15n + 2n for the 17n, and factor by grouping.

2[10n^2 + 17n + 3]
2[10n^2 + 15n + 2n + 3] {Substitution}
2[(10n^2 + 15n) + (2n + 3)] {Regrouping}
2[5n(2n + 3) + 1(2n + 3)] {Factoring each group}
2[(5n + 1)(2n + 3)] {Combining like groups}

The factoriztion is 2(5n + 1)(2n + 3).

Answer 8

20n^2 + 34n + 6 --- 2(10n^2 + 17n +3) --- 2(5n+1)(2n+3)

2(5n+1)(2n+3)

Answer 9

okayokay...so Jeff U is not right..

so it's 2(5n+1)(2n+3) after u divide by 2 and do the
5n--------1| 2
2n--------3| 15
------------------
10n^2----3| 17n

Answer 10

(2n+3)(10n+2)