Harvey is studying the polynomial f(x) = (2x - 3)(x - 4)(x2 + 3) and determines that the polynomial has four real roots. Is Harvey's statement CORRECT?
A. No, the polynomial f(x) has 2 real roots and 2 complex roots when solved.
O B. Yes, the polynomial f(x) has 4 x-intercepts so the function has 4 real roots.
C. Yes, the polynomial f(x) has a degree of 4 so the function has 4 real roots.
OD. No, the polynomial f(x) has 3 real roots because there are 3 factors.

It is A, I highly suggest you should read my explanation.

Step-by-step explanation:

(2x - 3)(x - 4)(x^2 + 3)

Exponent of 2x = 1

Exponent of x = 1

Exponent of x^2 = 2

Add them all together to get the degree. 1 + 1 + 2 = 4

Now we can tell that the polynomial can have 1,2,3,4, or no roots at all. The max number roots it can have is 4

To see the number of roots it has, we have to set each part equal to 0

2x - 3 = 0

Add 3 to both sides

2x - 3 + 3 = 0 + 3

2x = 3

Divide both sides by 2

2x/2 = 3/2

x = 3/2 or 1.5

Next,

x - 4 = 0

Add four to both sides

x - 4 + 4 = 0 + 4

x = 4

Then,

x^2 + 3 = 0

Subtract 3 from both sides

x^2 + 3 - 3 = 0 - 3

x^2 = -3

Find the square root of both sides

sqrt(x^2) = sqrt(-3)

x = i*sqrt(3), -i*sqrt(3)

Those are imaginary numbers, therefore there are not real.

The real roots are x = 3/2, 4

Harvey stated that this equation has four real roots. We can cross options B and C out since they stated he was correct.

It is not D because finding a square root of a number gives two answers; the positive and negative version, so it will have two solutions. There were only two real roots and the other two roots are complex roots.

That leaves us to option A, which states that there are 2 real and 2 complex roots. It is correct