Factor & Remainder Theorems - Algebra 3

1. When $P(x)$ is divided by $x-2$, the quotient is $x^2+3x-1$ and the remainder is 4. What is $P(3)$?
   Answer: $21$ From the division statement, $P(x)=(x-2)(x^2+3x-1)+4=x^3+x^2-7x+6$. Thus $P(3)=27+9-21+6=21$. Choice A is the remainder; choice C is the value of $(x-2)Q(x)$ at $x=3$ without adding 4; choice B is a common expansion error.
2. Which of the following is the complete factorization of $g(x)=x^4-5x^3+6x^2+4x-8$ over the integers?
   Answer: $(x+1)(x-2)^3$ When you see a quartic polynomial to factor over the integers, your first move should be the Rational Root Theorem: test factors of the constant term, $-8$, so $\pm1,\pm2,\pm4,\pm8$. Here, $g(2)=0$ and $g(-1)=0$, so $x-2$ and $x+1$ are factors. Dividing $g(x)$ by $x-2$ gives $x^3-3x^2+4$, and dividing that cubic by $x-2$ again gives $x^2-x-2$. This quadratic factors as $(x-2)(x+1)$. Pulling it all together, $g(x)=(x+1)(x-2)^3$. That is the complete integer factorization. Now look at the wrong choices. $(x-1)(x+2)^3$ would require $x=1$ and $x=-2$ to be roots, but $g(1)=-2$ and $g(-2)\neq0$, so this is just a mismatch of roots. $(x+1)(x-2)^2(x+2)$ includes the correct $(x+1)$ and two $(x-2)$ factors, but the extra $(x+2)$ is not a factor, and you would be missing one $(x-2)$; checking $g(-2)\neq0$ rules it out. Finally, $(x-1)(x-2)^3$ gets the repeated $(x-2)$ right but replaces $(x+1)$ with $(x-1)$, and $g(1)\neq0$ proves $(x-1)$ is not a factor. Your takeaway: for factoring polynomials over the integers, systematically test rational roots and divide until you reach a quadratic. This prevents both missing repeated factors and inventing roots that don't actually satisfy the polynomial.
3. The polynomial $P(x)=x^4-6x^3+14x^2-14x+5$ has real coefficients and $2+i$ is a zero. Which of the following is the complete factorization of $P$ over the real numbers?
   Answer: $(x^2-4x+5)(x-1)^2$ Because the coefficients are real, $2-i$ is also a zero, so $x^2-4x+5$ is a factor. Dividing $P$ by $x^2-4x+5$ gives $x^2-2x+1=(x-1)^2$. Choice A has the wrong sign in the quadratic factor; choice B includes complex factors and the wrong real root; choice C uses $(x+1)^2$ instead of $(x-1)^2$.
4. Let $P$ be a polynomial. When $P(x)$ is divided by $x-1$, the remainder is 5, and $P(-2)=0$. What is the remainder when $P(x)$ is divided by $(x-1)(x+2)$?
   Answer: $\frac{5}{3}x+\frac{10}{3}$ The remainder must be linear, so write $R(x)=ax+b$. Since $P(1)=5$, $a+b=5$. Since $P(-2)=0$, $-2a+b=0$. Solving gives $a=\frac{5}{3}$ and $b=\frac{10}{3}$. Choice A has a sign error; choice C swaps the coefficients; choice D overlooks that the two values determine the linear remainder.
5. What is the remainder when $x^{100}-2x^{49}+4$ is divided by $x+1$?
   Answer: $7$ Whenever you see "remainder when a polynomial is divided by $x+1$," think of the Remainder Theorem: the remainder is just the polynomial evaluated at the root of the divisor. Since $x+1=0$ gives $x=-1$, you need to compute $P(-1)$ for $P(x)=x^{100}-2x^{49}+4$. Now evaluate carefully. $(-1)^{100}=1$, because 100 is even. $(-1)^{49}=-1$, because 49 is odd. So $$P(-1)=1-2(-1)+4=1+2+4=7.$$ That is the remainder, so the correct answer is $7$. The other choices come from common sign or evaluation slips. The choice $3$ is what you get if you mistakenly plug in $x=1$, since $1-2+4=3$; remember, $x+1$ means $x=-1$, not $x=1$. The choice $5$ likely comes from thinking $(-1)^{100}=-1$, which would give $-1+2+4=5$; but any even power of $-1$ is positive $1$. The choice $1$ may come from only handling the first term, or forgetting that $-2(-1)$ adds $2$ and then $+4$ adds another $4$. Your study tip: for division by $x-c$, evaluate the polynomial at $c$. For $x+1$, use $c=-1$. Always check the parity of the exponents — even powers of $-1$ are $1$, odd powers are $-1$. This will make remainder questions quick and reliable.
6. Let $P(x)=x^3+ax^2-5x+6$. If $x-3$ is a factor of $P$, what is the remainder when $P$ is divided by $x+1$?
   Answer: $8$ Since $x-3$ is a factor, $P(3)=0$, so $27+9a-15+6=0$, giving $a=-2$. Then $P(x)=x^3-2x^2-5x+6$, and the remainder when divided by $x+1$ is $P(-1)=8$. The value $-8$ is a sign error; $0$ would mean $x+1$ is a factor; $18$ is $P(3)$ before solving for $a$.
7. A cubic polynomial $p$ has zeros $-2$, $1$, and $4$. If $p(0)=16$, which of the following could be $p(x)$?
   Answer: $2(x+2)(x-1)(x-4)$ When you see a cubic with given zeros, remember each zero $r$ corresponds to a factor $(x-r)$. So zeros $-2,1,4$ mean the factored form must contain $(x+2)(x-1)(x-4)$. The multiplier in front is then determined by $p(0)$. The choice $2(x+2)(x-1)(x-4)$ has exactly the correct factors. Checking $x=0$: $2(2)(-1)(-4)=16$, which matches $p(0)=16$. So this is the correct polynomial. Now look at why the others fail. The choice $2(x-2)(x+1)(x-4)$ has zeros $2,-1,4$, not $-2,1,4$, so it's not even the right root set. The choice $-2(x+2)(x-1)(x-4)$ has the correct zeros, but the leading multiplier is $-2$, so $p(0)=-16$, not $16$. The choice $2(x+2)(x+1)(x-4)$ gives zeros $-2,-1,4$, missing the zero $1$ and incorrectly including $-1$. So each wrong choice either misidentifies a sign in a factor or misapplies the multiplier. A strong strategy: first write the factors straight from the zeros, then evaluate at $0$ to solve for the leading coefficient. Also watch signs carefully: a zero of $-2$ requires $(x+2)$, not $(x-2)$. This avoids both root errors and constant errors.
8. Suppose $x+1$ and $x-2$ are factors of $p(x)=x^3+ax^2+bx-6$. What is the value of $p(3)$?
   Answer: $24$ Whenever you see linear factors of a polynomial, use the factor theorem: if $x-c$ is a factor, then $p(c)=0$. So $p(-1)=0$ and $p(2)=0$. Since $p(x)$ is a monic cubic and already has two linear factors, the third factor must be $x+k$. Multiplying the constant terms gives $(1)(-2)(k)=-6$, so $-2k=-6$, meaning $k=3$. Thus $$p(x)=(x+1)(x-2)(x+3).$$ Now evaluate: $$p(3)=(3+1)(3-2)(3+3)=4\cdot 1\cdot 6=24.$$ The value $0$ would be correct only if $x-3$ were a factor, but the third factor is $x+3$, not $x-3$. The value $-24$ is a sign error: at $x=3$, all three factors are positive, so $p(3)$ must be positive; the negative constant term does not make the value negative. The value $6$ is just the constant term, or its absolute value, and it ignores the contributions of the cubic and quadratic terms when evaluating $x^3+2x^2-5x-6$. A good habit: once you know the factors, multiply them out or compare constants to find the missing factor before evaluating.
9. Let $P(x)=x^4+ax^3-7x^2+bx+6$. If $P(1)=0$ and $P(-2)=0$, which of the following is the complete factorization of $P$?
   Answer: $(x-1)(x+1)(x+2)(x-3)$ From $P(1)=0$, $a+b=0$. From $P(-2)=0$, $-6-8a-2b=0$, so $4a+b=-3$. Solving gives $a=-1$ and $b=1$. Then $P(x)=x^4-x^3-7x^2+x+6=(x-1)(x+1)(x+2)(x-3)$. The other choices have incorrect zero sets.
10. If polynomial $P$ has remainder $-6$ when divided by $x-2$, what is the remainder when $P(x)+3x^2$ is divided by $x-2$?
    Answer: $6$ When you see "remainder when divided by $x-2$," the key is the Remainder Theorem: the remainder equals the polynomial evaluated at $x=2$. So $P(2)=-6$ is given. For $P(x)+3x^2$, evaluate the whole expression at $x=2$: $$P(2)+3(2)^2 = -6 + (3\cdot 4) = -6 +12 = 6.$$ Thus the remainder is $6$. Now, the traps: the choice $-6$ treats the added $3x^2$ as if it doesn't change the remainder; that would only be true if it were divisible by $x-2$, but at $x=2$, $3x^2$ contributes $12$. The choice $12$ correctly evaluates $3x^2$, but then forgets to add $P(2)=-6$, leaving out the original remainder. The choice $-18$ might come from multiplying the original remainder by $3$, or from subtracting $12$ instead of adding it; either way, it reverses the needed sign or overweights the given information. A useful habit: anytime a question asks for a remainder after division by a linear term, plug the root into the whole expression. For sums like $P(x)+Q(x)$, just evaluate $P(a)+Q(a)$. And be careful to include both contributions—do not forget the given $P(a)$ value.
11. Which polynomial has $x+1$ as a factor but does NOT have $x-1$ as a factor?
    Answer: $x^3+x^2+x+1$ Whenever you see "has a factor" on a polynomial question, your first move should be the factor theorem: if $x-c$ is a factor, then plugging $c$ into the polynomial gives $0$. Here you need $P(-1)=0$ for $x+1$ to be a factor, and $P(1)\neq 0$ for $x-1$ not to be a factor. Test the polynomial $x^3+x^2+x+1$. At $x=-1$, you get $-1+1-1+1=0$, so $x+1$ is a factor. At $x=1$, you get $1+1+1+1=4$, not $0$, so $x-1$ is not a factor. That matches the requirement. Now look at the others. $x^3+x^2-x-1$ gives $0$ at both $-1$ and $1$, so it has both factors. The same is true for $x^3-x^2-x+1$, which also gives $0$ at both. The remaining choice, $x^3-x^2+x-1$, gives $-4$ at $-1$ and $0$ at $1$, so it has $x-1$ but not $x+1$ — the opposite of what is asked. The quick takeaway: on factor questions, don't start factoring the whole polynomial. Plug the candidate roots into the polynomial. A zero means the factor exists; a nonzero value means it doesn't. Checking $P(-1)$ and $P(1)$ alone will separate these choices instantly.
12. A monic quartic polynomial $P$ with real coefficients has zeros $1$, $-2$, and $3$, and $P(0)=-12$. What is the fourth zero?
    Answer: $2$ Write $P(x)=(x-1)(x+2)(x-3)(x-r)$. Then $P(0)=(-1)(2)(-3)(-r)=-6r=-12$, so $r=2$. Choice A is a sign error; choice B repeats a known zero; choice D would give $P(0)=-24$.
13. A cubic polynomial $P(x)=x^3+ax^2+bx+c$ has zeros $2$ and $-1$, and $P(1)=12$. Which of the following is the factored form of $P$?
    Answer: $(x-2)(x+1)(x-7)$ Since $2$ and $-1$ are zeros, $P(x)=(x-2)(x+1)(x-r)$. Then $P(1)=(-1)(2)(1-r)=-2(1-r)=12$, so $1-r=-6$ and $r=7$. Choice B gives $P(1)=-16$; choice C has zeros $-2,1,7$; choice D has zeros $2,1,-7$.
14. A student claims that if $P$ is a cubic polynomial with leading coefficient 1, $P(2)=0$, and $P(-3)=0$, then $P(x)=(x-2)(x+3)(x-5)$. Which statement correctly diagnoses the claim?
    Answer: The third factor could be any linear factor $x-r$, so the two zeros do not determine a unique cubic polynomial. The Factor Theorem guarantees that $x-2$ and $x+3$ are factors, but the remaining factor is $x-r$ for an arbitrary real number $r$. Without more information, the third zero cannot be identified. Choice A incorrectly assumes uniqueness; choice C makes a sign error; choice D misstates the Factor Theorem.
15. Let $P(x)=x^4+3x^3-3x^2-11x-6$. How many distinct real zeros does $P$ have?
    Answer: $3$ Factoring gives $P(x)=(x+1)^2(x+3)(x-2)$, so the real zeros are $-1$ (with multiplicity 2), $-3$, and $2$. Counted distinctly, that is 3 zeros. Counting $-1$ twice gives 4; missing one of the other zeros gives 1 or 2.
16. What is the remainder when $P(x)=x^{100}-3x^{50}+2$ is divided by $x^2-1$?
    Answer: $0$ Since the divisor is quadratic, the remainder is linear: $R(x)=ax+b$. By the Remainder Theorem, $P(1)=R(1)$ and $P(-1)=R(-1)$. Compute $P(1)=1^{100}-3(1)^{50}+2=0$ and $P(-1)=(-1)^{100}-3(-1)^{50}+2=0$. Thus $a+b=0$ and $-a+b=0$, so $a=b=0$, and the remainder is $0$. The value $2$ is $P(0)$, $-2$ is a sign-error value, and $2x-2$ satisfies only $R(1)=0$, not $R(-1)=0$.
17. If $2x+1$ is a factor of $p(x)=6x^3+ax^2-5x+2$, what is the value of $a$?
    Answer: $-15$ When you see "factor of a polynomial," you're being tested on the factor theorem: if $2x+1$ is a factor, then $2x+1=0$ gives a root, so $x=-\frac12$ must make the polynomial equal zero. Plug that into $p(x)$: $6\left(-\frac12\right)^3 + a\left(-\frac12\right)^2 -5\left(-\frac12\right)+2=0.$ Simplify carefully: $6\left(-\frac18\right)+a\left(\frac14\right)+\frac52+2=0$ $-\frac34+\frac{a}{4}+\frac92=0.$ Combine the numbers: $-\frac34+\frac92 = -\frac34+\frac{18}{4}=\frac{15}{4}$. So: $\frac{a}{4}+\frac{15}{4}=0 \Rightarrow a+15=0 \Rightarrow a=-15.$ Now, why the others are traps. **$15$** comes from solving $a+15=0$ incorrectly — forgetting the negative sign. **$-1$** might result from misreading the root as $x=-\frac13$ or fumbling the fraction arithmetic. **$1$** usually appears when you accidentally substitute $x=\frac12$ instead of $-\frac12$, or when you lose track of signs in the coefficients. All of them would fail the factor theorem: only $a=-15$ makes $p\left(-\frac12\right)=0$. Study tip: Whenever a linear factor like $cx+d$ is given, immediately solve for the root $x=-\frac dc$ and substitute it into the polynomial, setting it equal to zero. Then write out every fraction step — sign errors are the most common trap on this type of question. If you get a non-integer, don't panic; simplify methodically and you'll land on the right value.
18. Let $f(x)=x^4-3x^3+ax^2-7x+6$. If $x-2$ is a factor of $f(x)$, what is the quotient when $f(x)$ is divided by $x-2$?
    Answer: $x^3 - x^2 + 2x - 3$ When you see "$x-2$ is a factor," use the factor theorem: $f(2)$ must equal $0$. That tells you the unknown coefficient before you divide. Compute $f(2)=16-24+4a-14+6=4a-16$, so $4a-16=0$ and $a=4$. Thus $f(x)=x^4-3x^3+4x^2-7x+6$. Now divide by $x-2$ using synthetic division with $2$: bring down $1$; $2\cdot 1+(-3)=-1$; $2\cdot(-1)+4=2$; $2\cdot 2+(-7)=-3$; $2\cdot(-3)+6=0$. The quotient is built from coefficients $1,-1,2,-3$, giving $$x^3-x^2+2x-3.$$ The choices ending in $+3$ — $x^3-x^2+2x+3$ and $x^3-x^2-2x+3$ — treat the final term with the wrong sign: the remainder is $0$, so the last quotient term must be $-3$, not $+3$. The choice $x^3-x^2-2x-3$ has the wrong $x$-coefficient: the step $2\cdot(-1)+4$ gives $2$, not $-2$. A quick check: $(x-2)(x^3-x^2+2x-3)$ multiplies back to $x^4-3x^3+4x^2-7x+6$. On this exam, when a factor is given, first plug the root into $f$ to find any unknown coefficient, then use synthetic division. The coefficients of the quotient come directly from the bottom row.
19. Suppose $x-3$ is a factor of $p(x)$. Which of the following must be a factor of $p(2x+1)$?
    Answer: $x-1$ Whenever you're told that $x-3$ is a factor of $p(x)$, the Factor Theorem is your key: $x-c$ is a factor exactly when $p(c)=0$. So here you know $p(3)=0$. Now the polynomial in the question is $p(2x+1)$, so ask: what value of $x$ makes the input equal to $3$? Solve $2x+1=3$, which gives $x=1$. Therefore $p(2(1)+1)=p(3)=0$, so $x-1$ must be a factor of $p(2x+1)$. That is the correct choice. . The choice $x-3$ is a tempting trap: you might copy the original factor unchanged. But plugging $x=3$ into the composed expression gives $p(7)$, not $p(3)$, so it does not have to be a factor. The choice $x+1$ would require $x=-1$, making the input $-1$, and you're not guaranteed that $p(-1)=0$. Similarly, $x-2$ would require $x=2$, making the input $5$, and no information guarantees $p(5)=0$. Those could be factors only if $p$ happened to have other specific roots, but they are not forced by the given fact. . Strategy: when a factor of $p(x)$ is given and you're asked about $p(\text{linear expression})$, transform the known root through that expression: set the linear expression equal to the known root and solve for $x$. The root shifts, but the factor does not stay the same.
20. A degree-4 polynomial $f$ satisfies $f(2)=0$ and $f(-2)\neq0$. Which statement must be true?
    Answer: $x-2$ is a factor of $f(x)$, but $x+2$ is not. When you see a polynomial value like $f(2)=0$, your first thought should be the Factor Theorem: if $f(c)=0$, then $x-c$ is a factor. Here, $c=2$, so $x-2$ is definitely a factor. At the same time, $f(-2)\neq 0$ tells you that $x+2$ is not a factor. Therefore the statement that must be true is exactly: "$x-2$ is a factor of $f(x)$, but $x+2$ is not." The degree being 4 simply means there are two more factors you don't know about; it doesn't force either of the missing factors. Now look at the wrong answers. The claim that both $x-2$ and $x+2$ are factors cannot be true, because $f(-2)\neq 0$ directly rules out $x+2$ as a factor. Similarly, the statement that "$x+2$ is a factor, while $x^2-4$ is not" fails because the first part is false. Finally, "$x^2-4$ is a factor" is impossible, since $x^2-4=(x-2)(x+2)$; for that to be a factor, both $x-2$ and $x+2$ would have to be factors, and you already know $x+2$ is not. Your study tip: whenever a question gives you a value like $f(a)=0$ or $f(a)\neq 0$, immediately write down which factor is confirmed and which factor is ruled out. Then translate compound factors like $x^2-4$ back into their linear pieces before deciding whether they can be factors. This avoids the common trap of assuming a degree-4 polynomial must include every nice-looking factor.

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