I already asked this question,I just wanted to clarify the answer.
Problem: The polynomial f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=-5, f(2)=12, then what are the x-intercepts of the graph of f?
Steve: clearly, f(0) = 0
so,
f(x) = ax^3+bx^2+cx
= x(ax^2+bx+c)
-1(a-b+c) = 15
1(a+b+c) = -5
2(4a+2b+c) = 12
solve those and you wind up with
f(x) = 2x^3+5x^2-12x
Now just solve for the roots of f(x)
How do I solve for the roots?
c'mon, it's just a quadratic:
x(2x^2+5x-12) = 0
x(2x-3)(x+4) = 0
...
To solve for the roots of a polynomial, in this case, the polynomial f(x) = 2x^3 + 5x^2 -12x, you can use the concept of factoring.
Step 1: Rewrite the polynomial equation as f(x) = 0.
In this case, f(x) = 2x^3 + 5x^2 -12x = 0.
Step 2: Factor out common terms from the equation, if possible.
In this case, the common term is x, so you can factor out x as follows:
x(2x^2 + 5x - 12) = 0.
Step 3: Solve for the roots by setting each factor equal to zero and solving for x.
Setting x = 0, you have one root: x = 0.
Now, for the quadratic factor, 2x^2 + 5x - 12 = 0, you can solve it using the quadratic formula or factoring.
Method 1: Quadratic Formula
The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients of the quadratic equation.
In this case, a = 2, b = 5, and c = -12.
Using the quadratic formula, you get:
x = (-5 ± √(5^2 - 4(2)(-12))) / (2(2))
x = (-5 ± √(25 + 96)) / 4
x = (-5 ± √121) / 4
x = (-5 ± 11) / 4
So, there are two additional roots: x = (6/4) = 3/2 and x = (-16/4) = -4.
Hence, the x-intercepts of the graph of f are at x = 0, x = 3/2, and x = -4.