Polynomial Division - Algebra 3

1. Use synthetic division to divide $2x^4 - 5x^3 + 3x - 7$ by $x - 3$. Which option gives the correct quotient and remainder?
   Answer: $2x^3 + x^2 + 3x + 12$, remainder $29$ Synthetic division with $c=3$ on the coefficients $2,-5,0,3,-7$ produces the bottom row $2,1,3,12,29$. Therefore the quotient is $2x^3+x^2+3x+12$ and the remainder is $29$. Using $c=-3$ instead gives $2x^3-11x^2+33x-96$ with remainder $281$; the variants with remainder $-29$ or $-3x$ in the quotient contain sign errors.
2. Given that $x=3$ is a root of $P(x)=2x^3-3x^2-11x+6$, which of the following is the complete factorization of $P(x)$?
   Answer: $(x-3)(2x-1)(x+2)$ Since $x=3$ is a root, the Factor Theorem tells you $x-3$ must be one factor. Use synthetic division with $3$ on the coefficients $2,-3,-11,6$: bring down $2$, multiply by $3$ to get $6$, add to $3$, multiply by $3$ to get $9$, add to $-2$, multiply by $3$ to get $-6$, add to $0$. The quotient is $2x^2+3x-2$, so $P(x)=(x-3)(2x^2+3x-2)$. But "complete factorization" means you must keep going: factor $2x^2+3x-2$ as $(2x-1)(x+2)$, since $(2x-1)(x+2)=2x^2+4x-x-2$. Therefore the complete factorization is $(x-3)(2x-1)(x+2)$. Why the others miss the mark? The choice $(x-3)(2x^2+3x-2)$ equals $P(x)$ but stopstoo early—it is a factorization, not the complete factorization. The choice $(x-3)(2x+1)(x-2)$ comes from sign errors in factoring the quadratic; expandingit gives $2x^3-9x^2+7x+6$, not $P(x)$. The choice $(x+3)(2x-1)(x+2)$ uses $x+3$ instead of $x-3$; since $3$ is a root, the factor must be $x-3$, and expanding that choice gives different signs altogether. Study tip: whenever you divide by a known root, immediately test whether the quotient factors further—especially a quadratic. If it does, "complete factorization" requires writing all linear factors. A quick multiplication check catches sign errors before you commit.
3. When $P(x)=x^3+kx^2-2x+4$ is divided by $x+1$, the remainder is $10$. What is $k$?
   Answer: $5$ Whenever you see a division-by-a-linear-factor question, the Remainder Theorem is your tool: the remainder when a polynomial $P(x)$ is divided by $x-a$ is $P(a)$. Here the divisor is $x+1$, which is $x-(-1)$, so you should evaluate $P(-1)$, not $P(1)$. Plug $x=-1$ into $P(x)=x^3+kx^2-2x+4$: $$P(-1)=(-1)^3+k(-1)^2-2(-1)+4=-1+k+2+4=k+5.$$ The problem says the remainder is $10$, so set $k+5=10$, giving $k=5$. Now let's see why the other choices are traps. Choosing $7$ usually comes from evaluating at $x=1$ instead of $x=-1$; that gives $P(1)=1+k-2+4=k+3$, and $k+3=10$ leads to $k=7$. Choosing $-5$ means you set $P(-1)=0$, treating the remainder as if the polynomial were divisible by $x+1$. Choosing $13$ often comes from a sign error on the constant term: if you accidentally compute $P(-1)=-1+k+2-4=k-3$, then $k-3=10$ gives $k=13$. Your takeaway: for a divisor like $x+1$, always use $x=-1$, and be careful with every sign when substituting. A quick synthetic division check can confirm your remainder before moving on.
4. Let $a$ be a real number. If $x^4 - ax^3 + 2x^2 + x - 4$ has remainder $6$ when divided by $x+1$, what is $a$?
   Answer: $8$ By the Remainder Theorem, the remainder is $P(-1)$. Since $P(-1)=1+a+2-1-4=a-2$, set $a-2=6$ to get $a=8$. Evaluating at $x=1$ instead gives $-a=6$, so $a=-6$; a sign error in the $-a(-1)^3$ term gives $a=-8$.
5. What is the remainder when $x^4-3x^3+x-2$ is divided by $x^2+1$?
   Answer: $4x-1$ When you see a polynomial divided by a quadratic like $x^2+1$, the remainder must have degree less than 2, so it will be of the form $ax+b$. A fast way to find it is to use the divisor as a simplification rule: since $x^2+1=0$, we have $x^2=-1$. Then $x^4=(x^2)^2=(-1)^2=1$, and $x^3=x\cdot x^2=-x$. Substituting into the dividend gives $$1-3(-x)+x-2=1+3x+x-2=4x-1.$$ So the remainder is $4x-1$. Each wrong choice comes from a small substitution or sign error. $4x+1$ uses the correct $1+3x+x$ but forgets to include the constant $-2$. $-4x-1$ makes two sign errors: treating $x^3$ as $x$ instead of $-x$, and treating the original $+x$ term as $-x$, while keeping the $-2$: $$1-3x-x-2=-4x-1.$$ $-4x+1$ uses those same two sign errors but drops the $-2$, giving $$1-3x-x=-4x+1.$$ So the trap here is not the polynomial division itself, but careless sign and constant tracking during substitution. Study tip: whenever the divisor is $x^2+1$, replace $x^2$ with $-1$, reduce powers, and simplify carefully. You can always check your remainder by substituting $x=i$ and verifying it matches $ai+b$.
6. When $x^3+2x^2-7x+k$ is divided by $x-2$, the quotient is $x^2+4x+1$ and the remainder is zero. What is $k$?
   Answer: $-2$ Whenever a division problem gives you the quotient and the remainder, remember the core relationship: dividend = divisor × quotient + remainder. Here the remainder is zero, so multiply $x-2$ by $x^2+4x+1$ to reconstruct the original polynomial. $$(x-2)(x^2+4x+1)=x^3+4x^2+x-2x^2-8x-2=x^3+2x^2-7x-2$$ Now compare that result to $x^3+2x^2-7x+k$. The constant terms must match, so $k=-2$. The wrong choices each come from a common slip. Choosing 2 means you forgot the negative sign in $-2\cdot 1$, turning the product's constant into $+2$. Choosing -1 often happens in synthetic division: the last quotient coefficient is $1$, and if you mistake that for the remainder, you solve $k+2=1$ and get $-1$. Choosing 1 is just grabbing the constant term from the quotient, but $k$ is the dividend's constant, which only appears after you multiply the divisor and quotient together. Your quickest check: set $x=2$ in the original polynomial and require the result to be $0$. That gives $8+8-14+k=0$, so $k=-2$. On test day, whenever you see a division setup, multiply the divisor and quotient to find the dividend—then compare term by term, especially the constant.
7. Suppose $x^3-2x^2+ax+b$ is divided by $x^2-x-1$, with quotient $x-1$ and remainder $2x+3$. What are $a$ and $b$?
   Answer: $a=2$, $b=4$ Using $P=D\cdot Q+R$: $(x^2-x-1)(x-1)=x^3-2x^2+1$. Adding the remainder $2x+3$ gives $x^3-2x^2+2x+4$. Therefore $a=2$ and $b=4$. The pair $a=0,b=1$ is the product before the remainder is added, and the other pairs come from sign errors in the multiplication.
8. If $P(x)$ is divided by $x-2$, the quotient is $Q(x)$ and the remainder is $5$. What is the remainder when $xP(x)$ is divided by $x-2$?
   Answer: $10$ When you see a divisor like $x-2$, the Remainder Theorem should come to mind immediately: the remainder when a polynomial is divided by $x-a$ is just the polynomial's value at $x=a$. Here, since $P(x)$ divided by $x-2$ leaves remainder $5$, that tells you directly that $P(2)=5$. Now you are asked about $xP(x)$ divided by $x-2$. The same theorem applies: evaluate $xP(x)$ at $x=2$. That gives $2 \cdot P(2) = 2 \cdot 5 = 10$. So the remainder is $10$. Why the others are traps: choosing $5$ means you forgot the extra factor of $x$ in front of $P(x)$ — the remainder changes because you're evaluating $xP(x)$, not $P(x)$. Choosing $2$ confuses the remainder with the value of $x$ itself at $x=2$, but the remainder is the whole expression's value, not just the divisor's root. Choosing $0$ might come from thinking multiplying by $x$ somehow makes the polynomial divisible by $x-2$, but since $xP(2)=10$, not $0$, there is definitely a nonzero remainder. Study tip: whenever you multiply a polynomial by a simple factor, use the Remainder Theorem directly — substitute the root into the entire new expression. This avoids redoing polynomial long division entirely.
9. A student uses synthetic division to divide $6x^3 - 7x^2 + 8x - 3$ by $2x - 1$. The student uses $c = \frac{1}{2}$ and obtains the bottom row $6,\,-4,\,6,\,0$. If the student reports the quotient as $6x^2 - 4x + 6$, which is the correct result of the division?
   Answer: $3x^2 - 2x + 3$, remainder $0$ Synthetic division with $c=\frac12$ divides by $x-\frac12$, not by $2x-1$. Since $x-\frac12=\frac12(2x-1)$, the quotient coefficients obtained from synthetic division must be divided by $2$: $6,-4,6$ become $3,-2,3$. Thus the quotient is $3x^2-2x+3$ and the remainder is $0$. The reported quotient is the quotient for division by $x-\frac12$, and doubling the coefficients is the inverse error.
10. When $P(x)$ is divided by $x^2-2x+1$, the quotient is $x^2+2x-1$ and the remainder is $3$. What is $P(2)$?
    Answer: $10$ The Division Algorithm gives $P(2)=(2^2-2(2)+1)(2^2+2(2)-1)+3$. Since $2^2-2(2)+1=1$ and $2^2+2(2)-1=7$, $P(2)=1\cdot7+3=10$. Using only the quotient value gives $7$, and using only the remainder gives $3$.
11. What is the result when $2x^4 - 5x^3 + 3x^2 - 7x + 10$ is divided by $x+1$?
    Answer: $2x^3-7x^2+10x-17$, remainder $27$ Whenever you divide a polynomial by a linear divisor like $x+1$, the key is to set the divisor equal to zero: $x+1=0$, so $x=-1$. Use synthetic division with $-1$. Write the coefficients of the dividend: $2, -5, 3, -7, 10$. Bring down the $2$; multiply by $-1$ to get $-2$,and add to $-5$,giving $-7$. Then multiply $-7$ by $-1$ to get $7$,add to $3$,giving $10$. Multiply $10$ by $-1$ to get $-10$,add to $-7$,giving $-17$. Finally, multiply $-17$ by $-1$ to get $17$,and add to $10$,giving remainder $27$. So the quotient is $2x^3-7x^2+10x-17$ and the remainder is $27$. Thistbe also matches the Remainder Theorem: substituting $x=-1$ into the original polynomial gives $27$. The choice with quotient $2x^3-7x^2+10x-17$ but remainder $-27$ flips the sign of the remainder; using the theorem hastily with $-1$ might lead to sign errors, but the true value is positive $27$. The choice with quotient $2x^3-3x^2+0x-7$ and remainder $3$ is what you would get if you divided by $x-1$ instead of $x+1$ – its root is $+1$, not $-1$. The choice withthe same quotient but remainder $-3$ repeats that divisor-sign confusion and then also miscomputes the remainder. Watch the sign in the divisor: $x+1$ means use $-1$, not $+1$. A good habit is to check your answer visually by rewriting: $$(x+1)(2x^3-7x^2+10x-17)+27$$ expands exactly to the original polynomial. Synthetic division with the correct root is faster and safer than long division – just be meticulous with signs at every step.
12. A polynomial $P(x)$ has remainder $5$ when divided by $x-2$ and remainder $-3$ when divided by $x+1$. What is the remainder when $P(x)$ is divided by $(x-2)(x+1)$?
    Answer: $\frac{8}{3}x-\frac{1}{3}$ When you see polynomial division tied to known values, think Remainder Theorem: $P(c)$ is the remainder when $P(x)$ is divided by $x-c$. Since $(x-2)(x+1)$ is degree 2, the remainder here must be linear: $R(x)=ax+b$. From the given information, $P(2)=5$ and $P(-1)=-3$. The remainder must match those outputs, so: $$2a+b=5,\qquad -a+b=-3.$$ Solving gives $a=\frac{8}{3}$ and $b=-\frac{1}{3}$, so the remainder is $\frac{8}{3}x-\frac{1}{3}$. The traps show common mistakes. $\frac{8}{3}x+\frac{1}{3}$ has the correct slope but the wrong intercept: at $x=-1$, it gives $-\frac{7}{3}$, not $-3$. $2x+1$ only fits $R(2)=5$; at $x=-1$, it gives $-1$, so it misses the second condition. $-2x+9$ also only fits $R(2)=5$, and it gives $11$ at $x=-1$, far from $-3$. Strategy: when a remainder question gives values at specific $x$'s, write the unknown remainder as a polynomial one degree less than the divisor, then plug those $x$-values into the remainder itself. That turns polynomial division into a simple system of equations.
13. When $x^4-2x^3+ax^2+bx-3$ is divided by $x^2-x-2$, the remainder is $5x+1$. What is $a+b$?
    Answer: $6$ Long division leaves the remainder $(a+b-1)x+(2a-1)$. Equating this to $5x+1$ gives $2a-1=1$, so $a=1$, and $a+b-1=5$, so $a+b=6$. The value $5$ is obtained if the $-1$ in the x-coefficient is ignored, and the smaller values come from later arithmetic errors.
14. What is the result when $3x^3 - 2x^2 + x - 7$ is divided by $x - 2$?
    Answer: $3x^2+4x+9$, remainder $11$ Whenever you see a polynomial divided by a binomial like $x - 2$, you can use synthetic division—it's faster and less error-prone than long division for this format. Write the coefficients of the dividend $3,-2,1,-7$, and bring down the leading 3 to start. Multiply by 2 to get 6, add to -2 to get 4. Multiply 4 by 2 to get 8, add to 1 to get 9. Multiply 9 by 2 to get 18, add to $-7$ to get 11. The quotient coefficientes are $3,4,9$, giving $3x^2 + 4x + \ ومن then 9$, and the final 11 is the remainder. This matches the correct choice: ** $3x^2+4x+9$,remainder 11. The choice with remainder $-11$** comes from a sign slip—forgetting that when you multiply and add, the subtrahend's sign flips correctly; you always add the product in synthetic division, so remainders rarely stay negative from careless arithmetic. The choices ** $3x^2-8x+17$ with remainders $\pm 41$** share common errors. The $-8x$ coefficient suggests misunderstanding the multiplication step: perhaps adding $6$ to $-2$ incorrectly as $-8$ (treating subtraction as us subtracting the product $6$ instead of adding it). The $+ 17$ coefficient follows from repeating that mistake, and the remainders $\pm 41$ result from carrying that error through and then mishandling the final sign. Your study takeaway: in synthetic division, keep a tidy row of coefficients, always add column-wise (never subtract), and write the final number as "remainder $r$" not as part of the quotient. One quick sign check at the end—plug $x=2$ into the original polynomial—should give $3(8)-2(4)+2-7 = 11$, confirming the remainder. Use that check to catch arithmetic slips before you choose.
15. Let $P(x)=2x^4+ax^3+bx^2-8x+3$. To check whether $P(x)$ is divisible by $(x-1)^2$, a student applies synthetic division twice with $c=1$. Which system of equations must be satisfied?
    Answer: $a+b=3$ and $3a+2b=0$ Dividing by $(x-1)^2$ requires two zero remainders when synthetic division by $x-1$ is done twice. For $P(x)=2x^4+ax^3+bx^2-8x+3$, the first synthetic division with $c=1$ gives the last entry $a+b-3$, so it must equal $0$: $a+b=3$. The quotient after this first division is $2x^3+(a+2)x^2+(a+b+2)x+(a+b-6)$. Applying synthetic division by $1$ to this quotient gives the last entry $3a+2b$, so the second required condition is $3a+2b=0$. Thus the correct system is $a+b=3$ and $3a+2b=0$. The distractor with $2a+3b=0$ reverses the coefficients in the second condition, and the options with $a+b=-3$ come from solving the first remainder equation with the wrong sign.
16. If $P(x)$ has remainder $3$ when divided by $x-2$, what is the remainder when $P(x)+x^2$ is divided by $x-2$?
    Answer: $7$ Whenever you see a question about remainders when dividing by a linear factor like $x-2$, think of the Remainder Theorem: the remainder is the value of the polynomial at $x=2$. So $P(2)=3$. Now for the new polynomial $P(x)+x^2$, the remainder when divided by $x-2$ is found by plugging in $2$ again: $$P(2)+2^2 = 3+4=7.$$ So the remainder is $7$. Why are the others traps? - **$3$** is the remainder of $P(x)$ alone. It forgets that the problem asks about $P(x)+x^2$, so the extra $x^2$ term must also be evaluated and added. - **$4$** is just the value of $x^2$ at $x=2$. It ignores $P(2)=3$ entirely. - **$1$** might come from subtracting $4-3$, but the polynomial is a sum, not a difference, so the values must be added. A good study takeaway: whenever you divide by $x-c$, the remainder is simply the polynomial's value at $c$. If the polynomial has several terms added together, just evaluate all of them at $c$ and add the results. This small idea appears often on algebra exams, so make it automatic.
17. When $2x^3+ax^2-3x+2$ is divided by $2x-1$, the remainder is $4$. What is the value of $a$?
    Answer: $13$ The Remainder Theorem applies at the zero of the divisor: evaluate $P$ at $x=\frac12$. $P(\frac12)=\frac14+a\cdot\frac14-\frac32+2=\frac{a+3}{4}$. Setting this equal to $4$ gives $a=13$. Evaluating at $x=1$ gives $a=3$, and incorrectly using $2P(\frac12)=4$ gives $a=5$.
18. The polynomial $P(x)$ has remainder $5$ when divided by $x-1$, and remainder $-2$ when divided by $x+2$. What is the remainder when $P(x)$ is divided by $(x-1)(x+2)$?
    Answer: $\frac{7}{3}x + \frac{8}{3}$ Write the remainder as $R(x)=ax+b$, since its degree must be less than $2$. The conditions give $R(1)=a+b=5$ and $R(-2)=-2a+b=-2$. Solving gives $a=\frac73$ and $b=\frac83$, so $R(x)=\frac73x+\frac83$. Sign errors in solving produce the variants with $-\frac83$ or with different coefficients.
19. When $P(x)$ is divided by $x^2-4$, the quotient is $Q(x)$ and the remainder is $3x-2$. Which statement must be true?
    Answer: $P(2)=4$ and $P(-2)=-8$ By the Division Algorithm, $P(x)=(x^2-4)Q(x)+3x-2$. Substituting $x=2$ and $x=-2$ makes the product $(x^2-4)Q(x)$ equal to $0$, so $P(2)=3(2)-2=4$ and $P(-2)=3(-2)-2=-8$. Ignoring the remainder gives $0$ for both, and using only the coefficient of the remainder gives $3$ and $-3$.
20. A student correctly performs synthetic division of a cubic polynomial $P(x)$ by $x+3$ and obtains the bottom row $2,-1,4,-5$. Which statement is true?
    Answer: $P(x)=(x+3)(2x^2-x+4)-5$ When you see synthetic division, recall that dividing by $x+3$ means the divisor is $x - (-3)$, so the number in the box is $-3$. The bottom row from synthetic division always gives the quotient coefficients followed by the remainder. Here, the last number $-5$ is the remainder, and the first three numbers $2, -1, 4$ are the coefficients of a quadratic quotient, so the quotient is $2x^2 - x + 4$. Thus $P(x) = (x+3)(2x^2 - x + 4) + (-5)$, which simplifies to $(x+3)(2x^2-x+4) - 5$. That matches the correct choice. Now check each distractor. The option with $+5$ instead of $-5$ misreads the remainder sign—the remainder is already $-5$, so you add $-5$, not $+5$. The option using $(x-3)$ incorrectly flips the sign of the divisor; since the divisor is $x+3$, the factor must be $x+3$. Finally, the option with $(2x^3 - x^2 + 4x - 5)$ treats the bottom row as a cubic quotient, but dividing a cubic by a linear factor yields a quadratic quotient—the bottom row's first three entries give a degree-2 quotient, not degree-3. For the exam, always identify the remainder as the last number in the bottom row, then write $P(x) = (\text{divisor})(\text{quotient}) + \text{remainder}$. Double-check the sign of the constant in the divisor—$x+3$ becomes $(x - (-3))$. And verify the quotient degree is one less than the dividend's degree.

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