How do you find, by factoring, f(x)=6x^2+x-14?

Question

I just need someone to walk me through it please.

The exact question is:

Find (by factoring) the x intercept or intercepts of the quadratic function: f(x) = 6x^2+x-14

Answer 1

this polynomial does not have a simple factoring scheme

if you solve this via the quadratic equation, you will get for solutions:

x=[-1+/-sqrt[1^2-4(6)(-14)]/2*6
=(-1-sqrt[337]/12) and (-1+sqrt[337]/12)

your factors would look like:

(x-{-1-sqrt[337]}/12)(x-{-1+sqrt[337]}/12)

so unless you wrote down the wrong coefficients, this one doesn't factor easily

Answer 2

Use the quadratic formula in the website listed to get the instances where f(x)=0.

For the quadratic formula, a=6, b=1 and c will equal -14. b^2-4ac is greater than zero so there will be zeros. Then when you have the roots you can write the equation in this form:

y=6*(x-root #1)*(x-root#2). Now you just have to find the roots by using the quadratic formula.

Answer 3

This equation is of form ax2+bx+c
a = 6 b = 1 c = -14
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-1 +/-sqrt(1^2-4(6)(-14)]/(2)(6)
discriminant is b^2-4ac =337
x=[-1 +sqrt(337)] / (2)(6)
x=[-1 -sqrt(337)] / (2)(6)
x=[-1+18.36] / 12
x=[-1-18.36] / 12
The roots are 1.446 and -1.613 (x intercepts)
The x intercepts are found by setting y=0.

Answer 4

You cannot factor f(x) using rational numbers.

You need the quadratic formula to get solution.