AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The polynomial function $ p $ is given by $ p(x) = (x + 3)(x^2 – 2x – 15) $. Which of the following describes the zeros of $ p $?

(A) $ p $ has exactly two distinct real zeros.
(B) $ p $ has exactly three distinct real zeros.
(C) $ p $ has exactly one distinct real zero and no non-real zeros.
(D) $ p $ has exactly one distinct real zero and two non-real zeros.

▶️ Answer/Explanation

Detailed solution

First, factor $ x^2 – 2x – 15 $:
$ x^2 – 2x – 15 = (x – 5)(x + 3) $.
So $ p(x) = (x + 3)(x – 5)(x + 3) = (x + 3)^2(x – 5) $.
Zeros: $ x = -3 $ (multiplicity 2), $ x = 5 $ (multiplicity 1).
Distinct real zeros: $ -3 $ and $ 5 $ (two distinct).
✅ Answer: (A)

Question

For the polynomial function $ g, \lim_{x \to -\infty} g(x) = -\infty $. Which of the following expressions could define $ g(x) $?

(A) $ -5x – 2x^4 $
(B) $ -5x^2 – 2x^7 $
(C) $ -9000x – \frac{x^5}{5} $
(D) $ -x^3 + 2x^4 $

▶️ Answer/Explanation

Detailed solution

End behavior is determined by the leading term (highest degree term). We need $ \lim_{x \to -\infty} g(x) = -\infty $. Check each leading term as $ x \to -\infty $:

(A) $ -2x^4 $ → degree 4 (even), coefficient negative → $ (-)\cdot(+\infty) = -\infty $ as $ x \to -\infty $ ✓
(B) $ -2x^7 $ → degree 7 (odd), coefficient negative → $ (-)\cdot(-\infty) = +\infty $ as $ x \to -\infty $ ✗
(C) $ -\frac{x^5}{5} $ → degree 5 (odd), coefficient negative → $ (-)\cdot(-\infty) = +\infty $ as $ x \to -\infty $ ✗
(D) $ 2x^4 $ → degree 4 (even), coefficient positive → $ +\infty $ as $ x \to -\infty $ ✗

Only (A) satisfies $ g(x) \to -\infty $ as $ x \to -\infty $.
✅ Answer: (A)

Question

The leading term of the polynomial function $ p $ is $ a_n x^n $, where $ a_n $ is a real number and $ n $ is a positive integer. The factors of $ p $ include $ (x – 3) $, $ (x – i) $, and $ (x – (2 + i)) $. What is the least possible value of $ n $?

(A) 3
(B) 4
(C) 5
(D) 6

▶️ Answer/Explanation

Detailed solution

Given factors: $ (x – 3) $ (real zero), $ (x – i) $ and $ (x – (2+i)) $ (nonreal).
For a polynomial with real coefficients, nonreal zeros occur in conjugate pairs: If $ i $ is a zero, so is $ -i $. If $ 2 + i $ is a zero, so is $ 2 – i $.
Thus least possible zeros: $ 3, i, -i, 2+i, 2-i $ → total 5 distinct zeros ⇒ degree at least 5.
✅ Answer: (C)

Question

The function $ Q $ is a polynomial of degree 3. If $ Q(5) = 0 $, which of the following must be true?

(A) $ Q(-5) = 0 $
(B) $ Q $ has two complex zeros.
(C) $ Q(x) $ can be expressed as $ (x – 5) \cdot P(x) $, where $ P(x) $ is a polynomial of degree 2.
(D) $ Q(x) $ can be expressed as $ \frac{P(x)}{x-5} $, where $ P(x) $ is a polynomial of degree 4.

▶️ Answer/Explanation

Detailed solution

$ Q(5) = 0 $ ⇒ $ x – 5 $ is a factor of $ Q $.
Since $ Q $ has degree 3, factoring out $ (x – 5) $ leaves a polynomial of degree 2: $ Q(x) = (x – 5) P(x) $, deg $ P $ = 2.
(A) not necessarily — only 5 is given as zero.
(B) not necessarily — the other two zeros could be real.
(D) incorrect — that would make $ Q $ degree 3 only if $ P $ degree 4 and division yields degree 3, but that’s not required from given info.
✅ Answer: (C)

Question

Which of the following statements must be true?

(A) The graph of $ k $ has three $ x $-intercepts.
(B) $-0.478 + 0.801i$ is a zero of $ k $.
(C) The equation $ k(x) = 0 $ has four real solutions.
(D) The graph of $ k $ is tangent to the $ x $-axis at $ x = 17.997 $.

▶️ Answer/Explanation

Detailed solution

From the answer: If a polynomial has real coefficients and a non-real zero $ a+bi $, its conjugate $ a-bi $ is also a zero.
Given that $ -0.478 – 0.801i $ is a zero of $ k $, then its conjugate $ -0.478 + 0.801i $ must also be a zero.
Thus (B) must be true.
The other options depend on full polynomial, so are not necessarily true.
✅ Answer: (B)

Question

The binomial theorem can be used to expand the polynomial function $p$, given by $p(x)=(x-3)^{5}$. What is the coefficient of the $x^{3}$ term in the expanded polynomial?

(A) $(-3)^3 \cdot 10$
(B) $(-3)^2 \cdot 10$
(C) $(-3)^3 \cdot 5$
(D) $(-3)^2 \cdot 5$

▶️ Answer/Explanation

Detailed solution

1. Identify General Term:
$\binom{n}{k} a^{n-k} b^k$ for $(a+b)^n$. Here $a=x, b=-3, n=5$.

2. Find Term with $x^3$:
We need $x^{5-k} = x^3 \implies 5-k=3 \implies k=2$.

3. Calculate Coefficient:
$\binom{5}{2} (-3)^2 = 10 \cdot (-3)^2$.

✅ Answer: (B)

Question

The polynomial function $ p $ is given by $ p(x) = (x + 3)(x^2 – 2x – 15) $. Which of the following describes the zeros of $ p $?

▶️ Answer/Explanation

Detailed solution

First, factor the quadratic $ x^2 – 2x – 15 $:
\[ x^2 – 2x – 15 = (x – 5)(x + 3) \]
Thus, \[ p(x) = (x + 3)(x – 5)(x + 3) = (x + 3)^2(x – 5) \]
Zeros occur when $ p(x) = 0 $:
• $ (x + 3)^2 = 0 $ gives $ x = -3 $ (multiplicity 2).
• $ x – 5 = 0 $ gives $ x = 5 $ (multiplicity 1).

So there are two distinct real zeros: $ x = -3 $ and $ x = 5 $.
Since $ p $ is degree 3, it has exactly 3 zeros in total counting multiplicity, but only two distinct real ones.
✅ Answer: (A)

Question

The polynomial function $ k $ is given by $ k(x) = ax^4 – bx^3 + 15 $, where $ a $ and $ b $ are nonzero real constants. Each of the zeros of $ k $ has multiplicity 1. In the $ xy $-plane, an $ x $-intercept of the graph of $ k $ is $ (17.997, 0) $. A zero of $ k $ is $-0.478 – 0.801i$. Which of the following statements must be true?

▶️ Answer/Explanation

Detailed solution

$ k $ is degree 4 with real coefficients. Non‑real zeros occur in conjugate pairs.
Given zero $-0.478 – 0.801i$ ⇒ its conjugate $-0.478 + 0.801i$ must also be a zero.
Thus two non‑real zeros and two real zeros (since total degree 4).
One real zero is given as $ x \approx 17.997 $, so there is another real zero.
All zeros have multiplicity 1 ⇒ graph crosses $x$-axis at each real zero, not tangent.
Thus (B) must be true.
✅ Answer: (B)

Question

Consider the function $g(x) = x^4 + 6x^3 + 5x^2 – 2x – 10$. If $-1 + i$ is a zero, then what is the sum of all the real roots of $g(x)$?

a. $-7$
b. $-4$
c. $-8$
d. $-1.5$

▶️ Answer/Explanation

Detailed solution

Since coefficients are real, the complex conjugate $-1 – i$ is also a zero.
The sum of all four roots is given by $-\frac{b}{a} = -\frac{6}{1} = -6$.
The sum of the complex roots is $(-1 + i) + (-1 – i) = -2$.
Let the real roots be $r_1$ and $r_2$.
Then, $(r_1 + r_2) + (-2) = -6$.
Solving for the sum of real roots: $r_1 + r_2 = -6 + 2 = -4$.
The correct option is b. $-4$.

Question

Find the $3^{\text{rd}}$ term in the expansion of $(y^2 + 4x)^4$.

a. $97y^4x^2$
b. $96y^4x^2$
c. $99y^4x^2$
d. $6y^4x^2$

▶️ Answer/Explanation

Detailed solution

Identify the general term formula: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
For the $3^{\text{rd}}$ term, set $r = 2$, $n = 4$, $a = y^2$, and $b = 4x$.
Calculate the binomial coefficient: $\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6$.
Substitute values: $T_3 = 6 \cdot (y^2)^{4-2} \cdot (4x)^2$.
Simplify exponents: $T_3 = 6 \cdot (y^2)^2 \cdot 16x^2$.
Multiply the constants: $6 \cdot 16 = 96$.
Final result: $96y^4x^2$.
The correct option is b.

AP Precalculus -1.5 Polynomial Functions and Complex Zeros- MCQ Exam Style Questions - Effective Fall 2023