Factor: 2x^2-x+14?

Question

Is it factorable??

Answer 1

the best way to tell if a quadratic is factorable is to take the discriminant (b^2 - 4ac) and if it is a perfect square it is factorable, if it is not then it is not
a = 2, b = -1, c = 14
b^2 - 4ac = (-1)^2 - 4(2)(14) = 1 - 112 = -111
since the discriminant is negative, this tells you that it is an imaginary root . 111 is not a perfect square, so it cannot be factored even using imaginary numbers.

No it is not factorable.

Answer 2

The roots are:
x = (1 - √(1 - 4*2*14))/4, (1 + √(1 - 4*2*14))/4
x = (1 - √(1 - 112))/4, (1 + √(1 - 112))/4
x = (1 - √-111)/4, (1 + √-111))/4

The factors, therefore are:
(x - (1 - √-111)/4)(x - (1 + √-111))/4
or
(x - 1/4 + (j/4)√111)(x - 1/4 - (j/4)√111)
where j = √-1

The expression is factorable, but the factors lie in the complex plane.

Answer 3

if this was written as ax^2 + bx + c then it will be factorable only if the amount (b^2 - 4ac) is positive or zero.
Calculating this amount, we get: 1-4*2*14 < 0. Since it's negative then the above is not factorable

Answer 4

Remember the quadratic formula? The part under the radical (b^2 - 4ac) is called the discriminant. If the discriminant is negative (as it is in your problem) then the polynomial has no real roots.

Use your graphing calculator to graph the polynomial and you will see it does not cross the x-axis. Therefore no roots, therefore not factorable.

Answer 5

No.
Because delta is negative
a=2
b=-1
c=14
delta=b*b-4ac
=1-4*2*14
=1-112
=-111<0
no roots

Answer 6

Hi,

No, this does not factor, as there are no factors of 28 that add to one. You would have to use the quadratic formula to solve this.

I hope this helps. :-)

Answer 7

This polynomial is already prime.
It does not factor.
none of the possibilities work.