Zeros & Multiplicity - Algebra 3

1. Let $P(x)=(x+1)^3(x-2)^2(x-4)$. Which statement correctly describes the sign of $P(x)$ on the intervals determined by its zeros?
   Answer: $P(x)>0$ on $(-\infty,-1)\cup(4,\infty)$ and $P(x)<0$ on $(-1,2)\cup(2,4)$. The zeros are $-1$ with odd multiplicity, $2$ with even multiplicity, and $4$ with odd multiplicity. Since the leading coefficient is positive, testing $x=-2,0,3,5$ gives signs $+, -, -, +$. Thus $P(x)>0$ on $(-\infty,-1)\cup(4,\infty)$ and $P(x)<0$ on $(-1,2)\cup(2,4)$.
2. For what value of $k$ does $P(x)=x^4-5x^3+4x^2+kx-16$ have $x=2$ as a zero of multiplicity 2?
   Answer: $12$ For $x=2$ to have multiplicity 2, $P(x)$ must be divisible by $(x-2)^2=x^2-4x+4$. Dividing $P(x)$ by $x^2-4x+4$ leaves remainder $(k-12)x$, so $k=12$. Then $P(x)=(x-2)^2(x^2-x-4)$, and $x^2-x-4$ is not zero at $x=2$, so the multiplicity is exactly 2.
3. A polynomial with real coefficients has zeros $0$ (multiplicity 1), $-3$ (multiplicity 2), and $5i$ (multiplicity 1). What is the minimum possible degree of the polynomial?
   Answer: $5$ A real-coefficient polynomial must include the conjugate $-5i$ as a zero too; summing multiplicities gives $1+2+1+1=5$.
4. A student claims: There is a degree-5 polynomial with real coefficients whose zeros, including multiplicity, are $-2$ (multiplicity 2), $3i$, and $4-i$. Which is the best reason this claim is false?
   Answer: The conjugates $-3i$ and $4+i$ are also forced, giving six zeros counting multiplicity. Whenever you see zeros of a polynomial with real coefficients, immediately think about conjugate pairs. Nonreal zeros cannot appear alone: if $a+bi$ is a zero, then $a-bi$ must also be a zero. Here the listed zeros are $-2$ (multiplicity 2), $3i$, and $4-i$. Since the polynomial must have real coefficients, $3i$ forces $-3i$, and $4-i$ forces $4+i$. That gives: $-2$ counted twice, then $3i, -3i, 4-i, 4+i$ — six zeros counting multiplicity. A degree-5 polynomial cannot have six zeros counting multiplicity, so the claim is false. That is exactly the best reason. The other choices miss the real issue. "A polynomial with real coefficients cannot have two different nonreal zeros" is false — it can, as long as they come in conjugate pairs. "Every nonreal zero must have multiplicity 2; therefore the degree would be at least 8" is also false: nonreal zeros have multiplicity 1 in this situation, and the forced conjugates already bring the count to six, not eight. "A degree-5 polynomial with real coefficients must have exactly one real zero" is false as well — a degree-5 real polynomial can have three real zeros and two nonreal conjugates; the real zeros just have to sum in multiplicity to an odd number. Your study takeaway: whenever a polynomial has real coefficients, count every nonreal zero together with its conjugate. That check alone would catch this claim immediately.
5. A degree-6 polynomial with real coefficients has exactly two distinct real zeros, one of which has multiplicity 3, and $2-i$ is also a zero. What is the multiplicity of $2+i$?
   Answer: $1$ Let the other real zero have multiplicity $m$, and let the conjugate pair $2\pm i$ have multiplicity $n$. Then $3+m+2n=6$. Since $2-i$ is a zero, $n\ge 1$, so $m+2n=3$ forces $n=1$ and $m=1$. Thus $2+i$ has multiplicity 1.
6. Given $p(x)=x^5+x^4-x^3+x^2+4x+2$, what is the multiplicity of the zero $x=-1$?
   Answer: 3 Whenever a question asks for the multiplicity of a zero, you are really asking: how many times does the factor $(x+1)$ appear in the factorization of $p(x)$? A zero of multiplicity $m$ means $(x+1)^m$ divides $p(x)$, but $(x+1)^{m+1}$ does not. Start by testing $x=-1$: $$p(-1)= -1+1+1+1-4+2=0,$$ so $x=-1$ is a zero. Now divide $p(x)$ by $x+1$ using synthetic division to get $x^4-x^2+2x+2$. This quotient is still zero at $x=-1$, so divide again: $(x+1)^2$ leaves $x^3-x^2+2$. That also vanishes at $x=-1$, and a third division by $x+1$ gives $x^2-2x+2$. At $x=-1$, this equals $1+2+2=5\neq 0$, so the factor stops after three copies. Thus the multiplicity is 3. If you chose 1, you likely confirmed the zero but did not repeat the division to see how many times the factor occurs. Choosing 2 means you stopped after the second division even though $x^3-x^2+2$ still had $x=-1$ as a root. Choosing 4 would require $x^2-2x+2$ to be divisible by $x+1$, but its remainder at $x=-1$ is 5. Study tip: keep dividing by the same factor until the remainder is nonzero; the number of successful divisions is the multiplicity.
7. Which leading-coefficient-1 polynomial of least degree with real coefficients has $4$ as a zero of multiplicity 1 and $2-3i$ as a zero?
   Answer: $(x-4)(x^2-4x+13)$ Whenever you see a polynomial with real coefficients and a complex zero, the Complex Conjugate Root Theorem is your compass: if $a+bi$ is a zero, then $a-bi$ must also be a zero. Since $2-3i$ is given as a zero, $2+3i$ is forced to be one as well. The quadratic factor containing these two zeros is $(x - (2-3i))(x - (2+3i))$. Expanding gives $(x-2+3i)(x-2-3i) = (x-2)^2 - (3i)^2 = x^2 -4x +4 - (-9) = x^2 -4x +13$. Multiplying by the linear factor for the zero of multiplicity 1 at 4 yields $(x-4)(x^2-4x+13)$. This is the least-degree polynomial (degree 3) with real coefficients and leading coefficient 1. Now, check the traps. $(x-4)(x^2-4x-5)$ has a quadratic that factors to $(x-5)(x+1)$, giving real zeros, not complex ones. $(x-4)(x^2+4x+13)$ has the wrong sign on the $x$-term; its zeros are $-2 \pm 3i$, not $2 \pm 3i$. The option $(x-4)(x-2+3i)$ only includes one complex factor—it forgets the conjugate, so its expanded form would have complex coefficients, violating the requirement for real coefficients. For your exam, remember the shortcut: $(x - (a+bi))(x - (a-bi)) = x^2 - 2ax + (a^2 + b^2)$. This instantly gives you the correct quadratic, so you can avoid sign errors and missing conjugates. Always verify the middle term's sign and that complex roots come in pairs.
8. Which polynomial has a graph that touches the $x$-axis at $x=-2$, crosses the $x$-axis at $x=3$, and satisfies $P(0)<0$?
   Answer: $(x+2)^2(x-3)^3$ An even multiplicity at $-2$ makes the graph touch the axis, and an odd multiplicity at $3$ makes it cross. For $(x+2)^2(x-3)^3$, $P(0)=2^2(-3)^3=-108<0$. The negated version has $P(0)>0$. The expressions with odd multiplicity at $-2$ switch the required touch/cross behavior.
9. Given that $1+i$ is a zero of $P(x)=2x^4-9x^3+17x^2-16x+6$, which list gives all zeros of $P$?
   Answer: $1+i,\ 1-i,\ \frac{3}{2},\ 1$ Since the coefficients are real, $1-i$ is also a zero. Dividing $P(x)$ by $x^2-2x+2$ gives $2x^2-5x+3=(2x-3)(x-1)$. The zeros are $1+i$, $1-i$, $\frac{3}{2}$, and $1$.
10. A polynomial of least possible degree has real coefficients, touches the $x$-axis at $x=-1$, crosses at $x=2$, and satisfies $p(0)=6$. Which polynomial could be $p(x)$?
    Answer: $p(x)=-3(x+1)^2(x-2)$ When you see "touches" versus "crosses" at an x-intercept, think about multiplicity: touching means the factor has even multiplicity, crossing means odd multiplicity. Here, touching at $x=-1$ requires $(x+1)$ squared (or any even power), and crossing at $x=2$ requires $(x-2)$ to the first power (or odd). For the least possible degree, use the smallest exponents: $(x+1)^2(x-2)$. Now use $p(0)=6$: plugging in $0$ gives $a(1)^2(-2)=-2a=6$, so $a=-3$. Thus $p(x)=-3(x+1)^2(x-2)$. The choice $-3(x+1)(x-2)^2$ reverses the multiplicities—it touches at 2 and crosses at -1, which is the opposite of what you need. The choice $3(x+1)^2(x-2)$ has the correct shape but the wrong leading coefficient: $p(0)=3(1)^2(-2)=-6$, not 6. And $-3(x-1)^2(x+2)$ has zeros at $1$ and $-2$, not at $-1$ and $2$. Only the correct polynomial satisfies both the multiplicity conditions and the value at $0$. A quick strategy: first translate the graph behavior into factors with the right exponents, then use the given point to solve for the constant. Always check that the constant's sign matches the y-intercept you need.
11. Consider the polynomial $P(x)=x^4+8x^3+26x^2+40x+25$. Which of the following correctly identifies all real and complex zeros of $P$, along with their multiplicities?
    Answer: The zeros are $-2+i$ and $-2-i$, each with multiplicity 2. The polynomial equals $(x^2+4x+5)^2$. Solving $x^2+4x+5=0$ gives $x=-2\pm i$. Since the quadratic is squared, each root appears twice, so each has multiplicity 2. The roots are $-2+i$ and $-2-i$, each with multiplicity 2; the other choices use the wrong real part $2\pm i$ or wrong multiplicities.
12. A degree-5 polynomial with real coefficients has $1+i$ as a zero of multiplicity 2. What is the minimum number of distinct real zeros it must have?
    Answer: $1$ The conjugate $1-i$ must also have multiplicity 2, using 4 of the 5 degrees. A real polynomial of odd degree must have at least one real zero, and one is attainable, for example $(x^2-2x+2)^2(x-1)$.
13. A cubic polynomial $P$ with real coefficients has zeros $2$ and $1-i$, and $P(0)=4$. Which of the following is $P(x)$?
    Answer: $-x^3+4x^2-6x+4$ Real coefficients force $1+i$ to be the third zero. Thus $P(x)=a(x-2)(x^2-2x+2)$. Since $P(0)=-4a=4$, $a=-1$. Expanding gives $-x^3+4x^2-6x+4$.
14. Let $f(x)=(x-3)^4(x+2)^2$ and $g(x)=[f(x)]^3$. Which statement gives all zeros of $g$ with their multiplicities?
    Answer: $3$, multiplicity 12; $-2$, multiplicity 6 Whenever you see zeros and multiplicities, look at the factored form: each factor $(x-a)^k$ contributes a zero at $x=a$ with multiplicity $k$. Here, $f(x)=(x-3)^4(x+2)^2$, so $f$ has zero $x=3$ with multiplicity $4$, and $x=-2$ with multiplicity $2$. Then $g(x)=[f(x)]^3$ cubes the entire polynomial; cubing doesn't change where the zeros are, but it does multiply every exponent by $3$. Thus, $x=3$ gets multiplicity $4\cdot =12$,and $x=-2$ gets multiplicity $2\cdot3=6$. So the correct statement is $3$, multiplicity $12$; $-2$, multiplicity $6$. The choice saying $3$, multiplicity $4$; $-2$, multiplicity $2$ gives the zeros of $f$, not $g$, because it forgets to multiply by the exponent $3$. The choice saying $3$, multiplicity $12$; $2$, multiplicity $6$ catches the cubing but has the sign wrong: solving $x+2=0$ gives $x=-2$, not $x=2$. The choice saying $3$, multiplicity $4$; $2$, multiplicity $2$ has both mistakes: it gives $f$'s multiplicities and also flips the sign of the second zero. When you raise a factored polynomial to a power,take each original zero and multiply its multiplicity by that power. Also, always solve each factor equal to zero carefully — the sign is easy to reverse.
15. Which polynomial has exactly two distinct real zeros, exactly two nonreal zeros, and exactly one real zero of multiplicity 2?
    Answer: $(x-1)^2(x^2+4)$ Since $(x-1)^2(x^2+4)=(x-1)^2(x-2i)(x+2i)$, the real zero is $1$ with multiplicity 2, and the nonreal zeros are $2i$ and $-2i$. The expression $(x-1)(x+1)(x^2+4)$ has two distinct real zeros, but neither has multiplicity 2; $(x-1)^2(x-2)^2$ has no nonreal zero; and $(x-1)^3(x^2+4)$ has only one distinct real zero, with multiplicity 3.
16. A student claims that $P(x)=x^4+2x^2+1$ has four distinct nonreal zeros. Which statement correctly corrects the student?
    Answer: The zeros are $i$ and $-i$, each with multiplicity 2, so there are only two distinct nonreal zeros. The polynomial factors as $(x^2+1)^2$, so the zeros are $i$ and $-i$, each with multiplicity 2. There are only two distinct nonreal zeros, not four. The statement about being positive for every real $x$ only rules out real zeros, not complex zeros.
17. Given that $x=2$ is a zero of $p(x)=x^3-5x^2+8x-4$, which statement gives all zeros of $p$ with their multiplicities?
    Answer: $2$, multiplicity 2; $1$, multiplicity 1 Whenever you see a zero given for a polynomial, your first move should be to factor it using that zero — this is a direct application of the Factor Theorem. Since $x=2$ is a zero, $x-2$ is a factor. Divide $p(x)=x^3-5x^2+8x-4$ by $x-2$ using synthetic division: you get $x^2-3x+2$. That quadratic factors further as $(x-1)(x-2)$. So the full factorization is $$p(x)=(x-2)(x^2-3x+2)=(x-2)^2(x-1).$$ Counting linear factors tells you the zeros and their multiplicities: $x=2$ appears twice, so it has multiplicity 2, and $x=1$ appears once, so it has multiplicity 1. The correct pairing is therefore $2$, multiplicity 2; $1$, multiplicity 1. The statement that says $2$, multiplicity 1; $1$, multiplicity 2 reverses the two multiplicities — not supported by the factorization. The statement listing $-1$ instead of $1$ is a sign-error trap: $(x+1)$ would correspond to zero $x=-1$, and substituting $-1$ into $p(x)$ gives $-18$, not $0$. Likewise, the statement listing $-2$ with multiplicity 2 mistakes the factor $x-2$ for $x+2$; substituting $-2$ gives $-48$, not $0$. Study tip: once you have one zero, always divide and factor the quotient completely. The exponent on each linear factor is the multiplicity — that's the entire story.
18. Let $R(x)=\frac{(x-2)^2(x+3)}{(x-2)(x+3)^2}$. Which statement correctly describes the $x$-intercepts and the graph of $R$?
    Answer: There is no $x$-intercept; the graph has a hole at $x=2$ and a vertical asymptote at $x=-3$. When you see a rational function, factor everything and compare the numerator and denominator before simplifying. Here, $R(x)=\frac{(x-2)^2(x+3)}{(x-2)(x+3)^2}$ simplifies to $\frac{x-2}{x+3}$, but only where $x\neq 2$ and $x\neq -3$. The factor $(x-2)$ cancels completely, so $x=2$ is a hole: the function is undefined there, even though the simplified form would be $0$ at that point. Because the graph has no point at $x=2$, it does not cross the $x$-axis there—so there is no $x$-intercept. The remaining denominator factor $(x+3)$ is zero at $x=-3$, so that is a vertical asymptote: the graph increases or decreases without bound near it. Thus the correct description is: no $x$-intercept, a hole at $x=2$, anda vertical asymptote at $x=-3$. The wrong choices each come froma simplifying mistake. Saying "the $x$-intercept is $x=2$" confuses the value of the simplified expression with an actual point on the original graph; the hole removes that point. Saying"the graph has a hole at $x=-3$" mistakes the vertical asymptote for a hole—the $(x+3)$ factor does not cancel completely, so it remains in the denominator. Saying"the $x$-intercept is $x=-3$ and the vertical asymptote is $x=2$" reverses both roles: $x=-3$ makes the denominator zero, so it cannot be an intercept, and $x=2$ is cancelled, so it cannot be a vertical asymptote. Remember: after simplifying a rational function, any canceled factor creates a hole, whileany factor left in the denominator creates a vertical asymptote. An $x$-intercept must come from a numerator zero that is actually in the domain of the original function.
19. Let $f(x)=(x^2-4x+4)(x^2+6x+9)(x^2+1)$. Which of the following lists all zeros of $f$ with their multiplicities?
    Answer: $2$, multiplicity 2; $-3$, multiplicity 2; $i$ and $-i$, each multiplicity 1 When you see a polynomial given in factored form like this, your job is to read each factor as a statement about zeros. A factor of the form $(x - r)^k$ tells you that $r$ is a zero with multiplicity $k$. The factor $(x^2+1)$ is not a difference of squares over the reals, but over the complex numbers it factors as $(x - i)(x + i)$, giving zeros $i$ and $-i$, each with multiplicity $1$. Now examine each quadratic: $x^2 - 4x + 4 = (x - 2)^2$, so $2$ has multiplicity $2$. $x^2 + 6x + 9 = (x + 3)^2 = (x - (-3))^2$, so $-3$ has multiplicity $2$. The correct answer lists exactly these: $2$ multiplicity $2$, $-3$ multiplicity $2$, and $i, -i$ each multiplicity $1$. Now look at the traps. The choice that lists $-2$ and $3$ instead of $2$ and $-3$ confuses signs—check the constant terms: $x^2 - 4x + 4$ gives $+2$, not $-2$. The choice that omits the complex zeros forgets that $x^2+1$ has no real solutions but does have imaginary ones; the question asks for all zeros, not just real zeros. The choice that gives $2$ and $-3$ multiplicity $1$ ignores that the squares in the factors double those multiplicities. Your study tip: always factor completely, including complex factors when necessary, and count the exponent on each factor as the multiplicity. Watch for sign flips when reading zeros from $(x - r)^k$.
20. Which statement correctly describes the real zeros of $P(x)=x^5-6x^4+12x^3-8x^2$?
    Answer: $0$ has multiplicity 2 and $2$ has multiplicity 3. Factor as $x^2(x^3-6x^2+12x-8)=x^2(x-2)^3$. Therefore the real zeros are $0$ with multiplicity 2 and $2$ with multiplicity 3.

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