Question

how do you find the zeros h(x)=3x^2+13+4 by factoring

Answer 1

That's \[h(x) = 3x^2 + 13x + 4\] right?

Answer 2

yes.

Answer 3

To get that 3x^2 term, you need to multiply x and 3x. So our factors will look like
(3x + a)(x + b) for some value of a and b.

\[(3x+a)(x+b) = 3x^2 + 3bx + ax + ab \]
But we want that to be equal to our h(x)
\[3x^2 + 3bx + ax + ab = 3x^2 + (3b+a)x + ab = 3x^2 + 13x +4\]
That means that (3b+a) = 13, and ab = 4. Can you think of two numbers that make that true?

The zeros of h(x) = (x + a)(x+b)(x+c) etc. are the values where (x+a) = 0, (x+b) = 0 etc. because multiplying by 0 will make the whole polynomial = 0.

Answer 4

i have to find the zeros by factoring and first setting it to zero like
3x^2+13+4=0. i need factors of 4 that add to 13

Answer 5

I just told you how to factor it...and once you have it factored, how to find the roots.

Answer 6

You don't need factors of 4 that add to 13, you need factors of 4 that when one of them is multiplied by 3 and added to the other it add to 13

Answer 7

i have to find the zeros by factoring and first setting it to zero like
3x^2+13+4=0.
>> i need factors of 4 that add to 13
that is true if the number in front of x^2 is 1. Unfortunately, it is 3
so it is more complicated (see whpalm's posts)
now you need one of the factors of 4 *3 plus the other factor of 4 adds up to 13