Identifying Factors and Zeros of Polynomials
Factors and zeros are two views of the same information, but the sign flips between them, which the SAT tests directly.
Reading the concept isn't enough. The score comes from practicing the real Bluebook-style format and finding the exact mistakes you keep making.
What the SAT tests here
If $(x - r)$ is a factor, then $x = r$ is a zero. The factor $(x + 1)$ gives the zero $x = -1$; the sign reverses.
How to solve one, step by step
Example: zeros of $p(x) = (x + 1)(x - 4)$.
- $(x + 1) = 0 \Rightarrow x = -1$.
- $(x - 4) = 0 \Rightarrow x = 4$.
The mistakes that cost points
- Forgetting the sign flip. $(x + 1)$ is a factor, but the zero is $x = -1$.
- Ignoring multiplicity. A repeated factor affects how the graph behaves at that zero.
Practice questions
Try these the way you would on test day, then open the solution to check your method.
Easy
The function $f(x) = (x + k)(x - 2)$ has a zero at $x = 7$. What is the value of $k$?
- A$k = -2$
- B$k = 5$
- C$k = 7$
- D$k = -7$
Show solution
Answer: D, $k = -7$. Since $x = 7$ is a zero of $f$, substitute: $f(7) = (7 + k)(7 - 2) = 0$. This gives $(7 + k)(5) = 0$, so $7 + k = 0$, which means $k = -7$.
Medium
The polynomial $g(x) = (x - 1)^2(x + 4)$ is graphed in the $xy$-plane. Which statement correctly describes the graph's behavior at $x = 1$?
- AThe graph touches the $x$-axis at $x = 1$ because the $y$-value equals 1 at that point.
- BThe graph crosses the $x$-axis at $x = 1$ because it is a zero of the function.
- CThe graph touches the $x$-axis at $x = 1$ but does not cross it, because the factor $(x-1)$ has even multiplicity.
- DThe graph touches the $x$-axis at $x = -1$ but does not cross it.
Show solution
Answer: C, The graph touches the $x$-axis at $x = 1$ but does not cross it, because the factor $(x-1)$ has even multiplicity.. The zero $x = 1$ comes from the factor $(x-1)^2$, which has multiplicity 2 (even). When a zero has even multiplicity, the graph touches the $x$-axis and bounces back without crossing. The zero $x = -4$ has multiplicity 1 (odd), so the graph crosses there. Therefore the graph touches but does not cross at $x = 1$.
Hard
The polynomial $p(x) = (x - k)^3(x + 2)^2$ is graphed in the $xy$-plane. The graph crosses the $x$-axis at exactly one zero and touches (but does not cross) the $x$-axis at exactly one other zero. Given that one zero of $p$ is $x = 5$, which of the following must be true?
- A$k = 5$, but $x = 5$ is the touching zero because the exponent 3 is larger than 1.
- B$k = -5$, because the zero of $(x - k)^3$ is the opposite of $k$.
- C$k = 5$, because $x = 5$ is the crossing zero corresponding to the odd-multiplicity factor $(x - k)^3$.
- D$k = 5$, and the $y$-value of the graph is 5 at the crossing point.
Show solution
Answer: C, $k = 5$, because $x = 5$ is the crossing zero corresponding to the odd-multiplicity factor $(x - k)^3$.. A zero from a factor with odd multiplicity causes the graph to cross the $x$-axis; a zero from a factor with even multiplicity causes the graph to touch without crossing. The factor $(x-k)^3$ has odd multiplicity (3), so it produces a crossing zero. The factor $(x+2)^2$ has even multiplicity (2), so it produces a touching zero at $x = -2$. Since the crossing zero is $x = 5$, we need $(x - k)^3 = 0$ when $x = 5$, giving $k = 5$.
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