usethe rational zero theorem to find all real zeros of
the polynomisl function.Use the zeros to factor f over the real numbers.
f(x)=x^4+2x^33-7x^2-8x+12
what are the real zeros?>
use the real zeros to factor f
f(x)=
Try values of x = ±1, ±2, ±3, ±4, ±6
hint: x = 1 and x = 2 worked for me
so (x-1) and (x-2) are factors
do either long algebraic division or synthetic division to find the other two factors.
To find the real zeros of the polynomial function f(x) = x^4 + 2x^3 - 7x^2 - 8x + 12, we can use the Rational Zero Theorem. The Rational Zero Theorem states that if a polynomial function has rational zeros (zeros as fractions), then they must be in the form of p/q, where p is a factor of the constant term (12) and q is a factor of the leading coefficient (1 in this case).
Step 1: Find the factors of the constant term (12): ±1, ±2, ±3, ±4, ±6, ±12.
Step 2: Find the factors of the leading coefficient (1): ±1.
Step 3: Form all possible rational zeros as fractions: ±1, ±2, ±3, ±4, ±6, ±12.
Now, we can use these possible rational zeros to perform synthetic division or long division to see which values result in a zero remainder. We will start by testing the first possible rational zero.
Let's try x = 1:
Performing synthetic division or long division, we get the remainder = 0 when x = 1 is tested. So x = 1 is a zero of the polynomial.
Now, we have reduced the problem to a degree 3 polynomial. We can divide the original polynomial by (x - 1):
f(x) = (x - 1)(x^3 + 3x^2 - 4x - 12)
Now, we can continue using the Rational Zero Theorem with the reduced polynomial x^3 + 3x^2 - 4x - 12 to find the remaining zeros.
Repeat steps 1-3 with the reduced polynomial:
Factors of the constant term (-12): ±1, ±2, ±3, ±4, ±6, ±12.
Factors of the leading coefficient (1): ±1.
Possible rational zeros: ±1, ±2, ±3, ±4, ±6, ±12.
Testing x = -1, we find that the remainder is 0. So, x = -1 is another zero of the polynomial.
Now, we have reduced the problem to a degree 2 polynomial. Divide x^3 + 3x^2 - 4x - 12 by (x + 1), and we get:
f(x) = (x - 1)(x + 1)(x^2 + 2x - 12)
Factoring out the real zeros, we have:
f(x) = (x - 1)(x + 1)(x + 4)(x - 3)
Therefore, the real zeros of f(x) are x = -1, x = 1, x = -4, and x = 3. And the factored form of f(x) is f(x) = (x - 1)(x + 1)(x + 4)(x - 3).