Find all zeros of the following polynomial. Write the polynomial in factored form. f(x)=x^3-3x^2+16x-48
I put: x^2(x-3)+16(x=3)
(x-3)(x^2+16)
For zeros:
x-3=0
x=0
**My teacher stated check the equation solution again. What is the value for x and hence what is the zero for the polynomial based on the first factor?
**Can someone help me?
You factored the cubic polynomial correctly as
(x-3)(x^2 + 16)
The first term tells the function has a zero at x = 3.
The other roots are imaginary. x = +/- 4i
yes, I did put:
x^2+16=0
x^2= -16
x= + or -4i
But she wrote that on the first part that I submitted and I don't really know what it is that she wants me to do.
**So are you saying that the value for x is 3, but then what is the zero for the polynomial based on the first factor??
x-3 = 0
means x = 0
so the polynomial is zero at x = 3
also at - 4 i and at + 4 i
To find the zeros of a polynomial, you need to solve the equation f(x) = 0. In this case, the polynomial is f(x) = x^3 - 3x^2 + 16x - 48.
Your attempt to start factoring the polynomial is correct. Let's continue with factoring it completely and then find the zeros:
f(x) = (x - 3)(x^2 + 16)
To find the zeros, we set each factor equal to zero and solve for x:
1) x - 3 = 0 --> x = 3
2) x^2 + 16 = 0
To solve the second equation, we need to find the square root of -16. However, the square root of a negative number is not a real number. Therefore, the equation has no real solutions.
So, the only zero of the polynomial f(x) is x = 3.
However, it seems that you made a mistake in your initial attempt. You wrote "(x-3)(x^2+16)" as the factored form. The correct factored form is "(x - 3)(x^2 + 16)". Therefore, the polynomial cannot be factored any further.
To avoid confusion, make sure to double-check your calculations and be careful when writing equations and factoring polynomials.