how do you factor 4x2-x+3=0

Question

factor equation using zero property

Answer 1

The zero property says that if a is a "zero" of a polynomial p(x) (that is, a solution to the equation p(x) = 0), then x-a is a factor of the polynomial. 

 

Since you were asked to factor it using the zero property, you have to solve the equation in order to find the factors.

 

Let's suppose that the way you wrote the equation in the problem was not correct and should have been 4x² + x - 3 = 0 

 

Then use the quadratic formula to solve the equation:

 

x = (-b ± √(b²-4ac)) / 2a =

 

(-1 ± √(1 - 4*4*(-3))) / 8=

(-1 ± √((1-(-48))) / 8 =

(-1 ± √49) / 8 =

(-1 ± 7) / 8 =

-1 or 3/4

 

That means that x + 1 and x - 3/4 are both factors of the polynomial

 

If you multiply these together you get x² + x/4 -3/4

 

This is not the same as the original polynomial, but when using the zero rule, you can, and may have to, multiply the resulting polynomial by a constant to get to the original polynomial.

 

Since the coefficient of x² in the original polynomial is 4, multiply by 4 to get 4x² + x - 3.  That means that the factoring of the original polynomial is 4(x+1)(x-3/4) = (x+1)(4x-3).

 

 

BUT...  If the way you wrote the equation was CORRECT (4x² - x + 3 = 0), then it works the same way -- except that this time you're going to get complex numbers as roots. Without going through the details...

 

x = (1 ± i√47) / 8

 

In this case, the factors would be (x - 1 - i√47) and (x - 1 + i√47) and again you have to scale it up so that the coefficient of the x² term is 4.  So the factoring could be written as:

 

4(x - 1 - i√47)(x - 1 + i√47)

 

 

 

 

 

Answer 2

I believe that you may be asked to factor 4x^2 + x - 3 = 0 rather than 4x^2 - x + 3 = 0. However, I will address each of these, just to be certain.

 

If we need to factor, 4x^2 + x - 3 = 0 I would change the left side to (x + 1) (4x - 3)

Next, if the product of 2 factors is zero, then one of the factors must be zero (The property of zero rule.)

 

If (x + 1) = 0 , then x = -1  If not, then (4x - 3) = 0 and x = 3/4 or 0.75 as a decimal. If you replace x with 0.75, then both sides equal zero. If you replace x with -1, then both sides are zero.

 

However, if the original equation is actually 4x^2 - x + 3 = 0, then I claim that trinomial is prime (It can not be factored) We could use the quadratic formula to se if there is real number solution, but my guess is it will be imaginary.

I just did it. I obtained two complex roots 1/8 + 1/8 (square root of -47)  and  1/8 - 1/8 (square root of -47).

 

I wonder if the original problem may have been 4x^2 - x - 3 = 0. If so, then the left side factors out to (x -1) (4x +3).

In this case, the solution set would be {1, -0.75} instead of {-1, 0.75} as stated above.

 

No matter what, I am fairly certain the original problem is not meant to be 4x^2 = x + 3 = 0 since one could not factor it , and the solution contains no real roots.

Answer 3

4x² - x + 3 = 0

Knowing that the only 2 possibilities for the first term are either 4x * x or 2x * 2x you can experiment.

Knowing the only possibility for the last term is 3 * 1 makes this much easier.

Let's start with the 1st possibility for the 1st term for ease:

 

(4x (+/-) 3)(x (+/-) 1)

 

Now figure out if any combination can get to the second term:

 

4x - 3x = x ; Yes

 

Plug in the signs

 

(4x -3)(x + 1) = 0

 

Set both equations = 0 and solve then verify if one or the other is = 0

 

4x - 3 = 0

4x = 3

x = 3/4

 

x + 1 = 0 

x = -1

 

plug in both values one at a time to see if the original equation evaluates to 0 (they should both evaluate to 0):

 

easier one is -1

 

4(-1)² + (-1) - 3 = 0

4(1) -1 - 3 = 0

4 - 1 - 3 = 0

4 - 4 = 0

0 = 0 :  check

 

4(3/4)² + (3/4) - 3 = 0

4(9/16) + (3/4) - 3 = 0

(36/16) + (3/4) - 3 = 0

(36/16) + (12/16) - 3 = 0

(48/16) - 3 = 0

3 - 3 = 0

0 = 0 : check