How do you factor 3x^2-7x+20?

Answer 1

It's not possible, with any of the real combinations you can get. If you solve using the quadratic formula, you realize that the solutions are imaginary.

for ax^2+bx+c, quadratic = {-b[+-]sqrt(b^2-4ac)}/2a. In this case, a=3, b=-7, c=20

Plugging in the values for the quadratic equation, you get:

7 [+-] sqrt(49-240)/6. [the sqrt(-191) simplifies to i*sqrt(191)

Therefore your two solutions are [7+ i*sqrt(191)]/6 and [7 - i*sqrt(191)]/6

Thus, the factoring would be:
(x+[7+ i*sqrt(191)]/6)(x- [7 - i*sqrt(191)]/6).

I know its ugly, but that's what the solution is.

Hope this helps

Answer 2

Hi Dear Clashing ;

{ as you know if ;
f(x) = ax ² + bx + c
∆ = b ² - 4ac & ∆ ≥ 0
x = ( -b ± √∆) / 2a } ....

Step 1
f(x) = 3x ² - 7x + 20
if a = +3 , b = -7 , c = +20
∆ = ( 7 ) ² - 4 * ( 3) * (20)
∆ = 49 - 240 → ∆ = - 191 → ∆ < 0

so
This function has no real roots.

Good Luck Dear.

Answer 3

This doesn't factor nicely; in fact, it has no real roots.

If it was 3x^2 - 7x - 20, then you could look for a factorization of the form (3x+a)(x+b), where a+3b=-7 and ab=-20. a=4 and b=-5 would work in this case.

Answer 4

A quick way to check if there are any real solutions is to check the discriminant. It is defined as Δ=b^2 -4ac

If Δ<0, there are no real roots
If Δ=0, there are two REPEATED roots, ie/they are the same
ifΔ>0 there are two real solutions; furthermore, if the solution of Δ is a perfect square(2,4,9,16,25,36........), the roots are rational, where as if they aren't, the roots aren't rational.

So, if we find Δ for your equation, we get Δ=49-240<0, so there are no real roots. Complex roots are another problem.

Answer 5

This is no real root, since discriminant is less than 0. However, if we let i = SQR(-1), then,
7+-SQR(7x7-4x3x20) = SQR (191) i
so, the roots are = (7+SQR(191) i )/6 or (7-SQR(191) i )/6

Answer 6

It cannot be factorised