If f(x) = x^3 + x^2 - 24x + 36 and f(2) = 0, then find all of the zeros of f(x) algebraically.
Solution
We are tasked with solving the given cubic equation for its zeros, given f(x)=x3+x2−24x+36 and the condition f(2)=0. This implies x=2 is one of the roots. Let's proceed to find all the zeros algebraically:
Step 1: Use Synthetic or Long Division
Since x=2 is a root, (x−2) is a factor. Divide f(x) by (x−2).
Perform synthetic division:
Divide x3+x2−24x+36 by (x−2).
Coefficients: 1,1,−24,36
Synthetic division steps:
Multiply 2 by the previous result and add to the next coefficient.
Step through the process:
2 & 1 & 1 & -24 & 36 \\
& & 2 & 6 & -36 \\
\hline
& 1 & 3 & -18 & 0 \\
\end{array}$$
The quotient is $$x^2 + 3x - 18$$, and the remainder is $$0$$.
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### Step 2: Factor the Quadratic
We now have $$f(x) = (x - 2)(x^2 + 3x - 18)$$. Next, factor $$x^2 + 3x - 18$$.
1. Look for two numbers that multiply to $$-18$$ and add to $$3$$: $$6$$ and $$-3$$.
2. Factor the quadratic: $$x^2 + 3x - 18 = (x + 6)(x - 3)$$.
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### Step 3: Write the Complete Factorization
The fully factored form of $$f(x)$$ is:
$$f(x) = (x - 2)(x + 6)(x - 3).$$
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### Step 4: Identify the Zeros
Set $$f(x) = 0$$ and solve:
$$x - 2 = 0 \implies x = 2,$$
$$x + 6 = 0 \implies x = -6,$$
$$x - 3 = 0 \implies x = 3.$$
Thus, the zeros of $$f(x)$$ are:
$$\boxed{x = 2, x = -6, x = 3}.$$
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Would you like a detailed explanation for any specific step? Let me know!
Here are some related questions:
1. How do you verify that a root satisfies a polynomial equation?
2. What are some methods for factoring cubic polynomials?
3. How does synthetic division differ from long division of polynomials?
4. Can all cubic equations be factored algebraically?
5. How do you handle a polynomial with no rational roots?
**Tip**: Always check if a given root works by substituting it into the original polynomial!