Solving Polynomial Equations - Algebra 3

1. The polynomial $x^3+ax^2-7x+b=0$ has real coefficients and $2+i$ is one root. What is $a+b$?
   Answer: 14 Since the coefficients are real, $2-i$ is also a root. Their quadratic factor is $x^2-4x+5$. If the third root is $r$, the cubic is $(x^2-4x+5)(x-r)=x^3+(-4-r)x^2+(5+4r)x-5r$. Setting the $x$-coefficient equal to $-7$ gives $r=-3$. Hence $a=-1$ and $b=15$, so $a+b=14$. The choice $15$ is $b$, $-1$ is $a$, and $-7$ is the $x$-coefficient.
2. For what value of $k$ does $x^4-2x^2+k=0$ have exactly three distinct real solutions?
   Answer: 0 Let $u=x^2\ge0$. The equation becomes $u^2-2u+k=0$. A positive $u$ gives two real $x$-values, while $u=0$ gives one real $x$-value. To get exactly three distinct real solutions, one root must be $u=0$, so $k=0$. Then $u^2-2u=u(u-2)=0$, yielding $x=0, \pm\sqrt{2}$. For $k=-1$ or $k=1$ there are two distinct real solutions; for $k=2$ there are none.
3. A student is asked to list all roots of $x^3-3x^2+4=0$, counting multiplicities. The student tests $x=2$, divides by $x-2$, obtains $x^2-x-2$, and writes $2,-1$. Which correction is needed?
   Answer: The quotient factors as $(x-2)(x+1)$, so the roots are $2,2,-1$. The division gives $x^3-3x^2+4=(x-2)(x^2-x-2)$, and $x^2-x-2=(x-2)(x+1)$. Therefore the factored form is $(x-2)^2(x+1)$. Counted with multiplicity, the roots are $2,2,-1$. The student missed the second factor of $x-2$ in the quotient.
4. What is the sum of the squares of all distinct real solutions of $x^3 - 3x^2 - 4x + 12 = 0$?
   Answer: $17$ Whenever you see a cubic like this, try factoring by grouping before doing anything else. Group the first two terms and the last two terms: $x^3-3x^2-4x+12=x^2(x-3)-4(x-3)=(x^2-4)(x-3)=(x-2)(x+2)(x-3).$ So the distinct real solutions are $x=-2$, $x=2$, and $x=3$. The sum of their squares is: $(-2)^2+2^2+3^2=4+4+9=17.$ Thus $17$ is the correct choice. The other choices are traps from rushing. $3$ is simply the sum of the roots, using Vieta's formula from the coefficient of $x^2$, but the question asks for the sum of their squares, not the sum itself. $9$ is the square of that root-sum; it answers $\left(\sum roots\right)^2$, not $\sum roots^2$. $25$ often comes from squaring coefficients such as $(-3)^2+(-4)^2=25$ and treating coefficients as the solutions—but coefficients are not roots. A safe strategy: factor completely, list the actual distinct solutions, then square each one. If you prefer Vieta, remember: $r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_1r_3+r_2r_3).$ But here, factoring by grouping is the fastest and clearest path to the answer.
5. Let $P(x)=x^4-3x^3+ax^2+bx-6$. If $2$ and $-1$ are roots of $P(x)=0$, which list gives all roots of the equation?
   Answer: $-1, 2, 1+i\sqrt{2}, 1-i\sqrt{2}$ Using $P(2)=0$ gives $2a+b=7$, and using $P(-1)=0$ gives $a-b=2$. Thus $a=3$ and $b=1$. Dividing $x^4-3x^3+3x^2+x-6$ by $(x-2)(x+1)=x^2-x-2$ gives $x^2-2x+3$, whose roots are $1\pm i\sqrt{2}$. The full root list is $-1, 2, 1+i\sqrt{2}, 1-i\sqrt{2}$. The other choices use sign errors in the linear factors or in solving $x^2-2x+3=0$.
6. The roots of $x^3-6x^2+kx-6=0$ are positive integers, and the product of two of the roots is $3$. What is $k$?
   Answer: 11 For a monic cubic, the product of all roots is $-(-6)=6$. If one pair has product $3$, the remaining root is $6/3=2$. Substituting $x=2$ gives $8-24+2k-6=0$, so $2k=22$ and $k=11$. Indeed, $x^3-6x^2+11x-6=(x-1)(x-2)(x-3)$. The other choices come from using the constant term as $k$ or from incorrect Vieta pair sums.
7. A monic cubic polynomial $p(x)=x^3+ax^2+bx+c$ satisfies $p(1)=0$, $p(2)=0$, and $p(3)=2$. What is the value of $a$?
   Answer: $-5$ When you see a monic cubic like $x^3+ax^2+bx+c$ and are told values of the polynomial, your first move should be to use the known roots. Since $p(1)=0$ and $p(2)=0$, both $1$ and $2$ are roots, so the polynomial factors as $p(x)=(x-1)(x-2)(x-r)$ where $r$ is the third root. Expanding gives $p(x)=x^3-(r+3)x^2+(3r+2)x-2r,$ so $a=-(r+3)$. Now use $p(3)=2$: $(3-1)(3-2)(3-r)=2 \implies 2(1)(3-r)=2 \implies 3-r=1 \implies r=2.$ Thus the third root is also $2$, so $p(x)=(x-1)(x-2)^2=x^3-5x^2+8x-4,$ and $a=-5$. Why the others fail: choosing $-7$ would mean $r=4$, but then $p(3)=2(1)(-1)=-2$, not $2$. Choosing $-6$ would mean $r=3$, giving $p(3)=0$. Choosing $-4$ would mean $r=1$, giving $p(3)=2(2)(1)=4$, not $2$. Each of these mistakes comes from misidentifying the third root or plugging in incorrectly. Study tip: whenever a polynomial is monic and has given roots, write it in factored form first. Then use the remaining condition to solve for the unknown root before expanding. This avoids guessing among coefficients.
8. Given that $x=-3$ is a root of $x^3+kx^2-5x-15=0$, which list gives all roots of the equation?
   Answer: $-3, -\sqrt{5}, \sqrt{5}$ Substituting $x=-3$ gives $-27+9k+15-15=0$, so $k=3$. The polynomial becomes $x^3+3x^2-5x-15=x^2(x+3)-5(x+3)=(x+3)(x^2-5)$. Therefore the roots are $-3, -\sqrt{5}, \sqrt{5}$. The other choices either solve $x^2-5=0$ incorrectly, omit $-\sqrt{5}$, or use the wrong sign for the given root.
9. Which of the following is NOT a possible rational root of $6x^4 - 5x^3 + 2x - 7 = 0$?
   Answer: $\frac{5}{6}$ When you see a polynomial like $6x^4 - 5x^3 + 2x - 7 = 0$ and are asked about possible rational roots, the Rational Root Theorem is your tool: any rational root must be of the form $\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. Here, the constant term is $-7$, so $p$ can only be $\pm 1$ or $\pm 7$. The leading coefficient is $6$, so $q$ can be $\pm 1, \pm 2, \pm 3, \pm 6$. That means possible roots include $7$ (since $7 = \frac{7}{1}$), $\frac{7}{2}$ (since $7$ divides $7$ and $2$ divides $6$), and $\frac{1}{3}$ (since $1$ divides $7$ and $3$ divides $6$). But $\frac{5}{6}$ cannot be a rational root because its numerator, $5$, is not a factor of $7$. Even though the denominator $6$ is a factor of the leading coefficient, the numerator must divide the constant term — that's the trap this question sets. Remember to check both parts: the numerator must divide the constant, and the denominator must divide the leading coefficient. If either fails, that number is not a possible rational root. So when you see fractions, quickly test numerator against the constant term before doing any polynomial arithmetic.
10. The roots of $x^3-12x^2+kx-64=0$ are three positive real numbers in geometric progression. What is $k$?
    Answer: 48 Let the roots be $\frac{a}{r}, a, ar$. Their product is $a^3=64$, so the middle root is $a=4$. Their sum is $4(\frac{1}{r}+1+r)=12$, so $r+\frac{1}{r}=2$. Since $r>0$, this forces $r=1$, meaning the roots are $4,4,4$. Thus $k=4\cdot4+4\cdot4+4\cdot4=48$. The other choices confuse $k$ with one pairwise product or with the product of all roots.
11. Which set contains all real solutions of $4x^4 - 17x^2 + 4 = 0$?
    Answer: $\{-2, -\frac{1}{2}, \frac{1}{2}, 2\}$ Whenever you see an equation with only even powers of $x$—like $x^4$ and $x^2$—think "quadratic in disguise." Let $u = x^2$. Then the equation becomes $$4u^2 - 17u + 4 = 0.$$ Factor: $$(4u - 1)(u - 4) = 0.$$ So $u = \frac14$ or $u = 4$. Since $u = x^2$, solve each piece: $$x^2 = \frac14 \implies x = \pm \frac12,$$ $$x^2 = 4 \implies x = \pm 2.$$ Thus the full real solution set is $$\left\{-2, -\frac12, \frac12, 2\right\}.$$ The choice containing exactly these four values is the correct one. Now look at the wrong choices. The set $\{-2, 2\}$ misses the two solutions $-\frac12$ and $\frac12$; this happens if you only solve $x^2 = 4$ and forget that $x^2 = \frac14$ also gives real roots. The set $\{-\frac12, \frac12\}$ makes the opposite mistake: it includes only the smaller roots and misses $\pm 2$. The set $\{-2, -\frac14, \frac14, 2\}$ includes $\pm 2$ but has $\pm \frac14$ instead of $\pm \frac12$; this is a square-root error—if $x^2 = \frac14$, then $x = \pm \frac12$, not $\pm \frac14$. Your study tip: for even-powered equations, substitute $u = x^2$, solve for $u$, then take both positive and negative square roots. Always ask yourself, "Did I include both signs for every positive $u$?" That single habit prevents the most common trap on this type of question.
12. How many real solutions does $x^6-7x^3-8=0$ have?
    Answer: 2 Let $u=x^3$. Then $u^2-7u-8=(u-8)(u+1)$, so $x^3=8$ or $x^3=-1$. Each cubic has one real root and two nonreal roots, giving $2$ real solutions total. The choice $6$ counts all complex roots; $4$ and $3$ come from miscounting the roots of the two cubic factors.
13. A cubic polynomial has zeros $-2$, $1$, and $3$. If $f(0)=12$, what is $f(2)$?
    Answer: $-8$ When a cubic polynomial has known zeros, the most reliable move is to write it in factored form with an unknown leading coefficient: $f(x)=a(x+2)(x-1)(x-3).$ The zeros tell you the factors; the extra point determines $a$. Plug in $x=0$: $12=a(2)(-1)(-3)=6a,$ so $a=2$. Now evaluate at $x=2$: $f(2)=2(4)(1)(-1)=-8.$ Thus the correct value is $-8$. The distractor $-4$ is what you get if you evaluate $(x+2)(x-1)(x-3)$ at $x=2$ but forget to multiply by the leading coefficient $a=2$. The choice $4$ comes from the same kind of slip plus losing the negative sign: $(4)(1)(1)=4$. The choice $8$ arises when you remember to multiply by $a=2$ but accidentally treat $2-3$ as $1$ instead of $-1$, giving $2(4)(1)(1)=8$. Always start with factored form when zeros are given, then use the extra point to find the leading coefficient. Finally, evaluate each factor carefully at the target input — especially watch the sign of any negative factor like $(x-3)$.
14. One solution of the equation $x^4-8x^3+24x^2-32x+15=0$ is $x=2+i$. What is the complete set of real solutions?
    Answer: $x=1$ and $x=3$ only Since the coefficients are real, if $2+i$ is a root, then $2-i$ is also a root. Their quadratic factor is $(x-(2+i))(x-(2-i))=x^2-4x+5$. Dividing the quartic by this factor yields $x^2-4x+3=(x-1)(x-3)$. Therefore the only real solutions are $x=1$ and $x=3$. The choice $x=-1,-3$ comes from a sign error in solving the quotient; $x=2+i,2-i$ stops after the conjugate pair and misses the real roots from the quotient; and the four-value choice includes nonreal roots in addition to the real roots, so it is not a list of real solutions only.
15. Which equation has exactly the solutions $-1,\ 1+\sqrt{2},\ 1-\sqrt{2}$?
    Answer: $x^3-x^2-3x-1=0$ The quadratic factor from the two irrational roots is $(x-(1+\sqrt{2}))(x-(1-\sqrt{2}))=x^2-2x-1$. Multiplying by the factor for $-1$, $(x+1)(x^2-2x-1)=x^3-x^2-3x-1$. The other choices come from sign errors when expanding the conjugate factor or when multiplying by $x+1$.
16. An open-top box is made by cutting congruent squares of side length $x$ inches from each corner of a 10-inch by 12-inch rectangular sheet and folding up the sides. The box has volume 96 cubic inches. Which list gives all valid values of $x$?
    Answer: $2$ and $\frac{9-\sqrt{33}}{2}$ inches The volume equation is $x(10-2x)(12-2x)=96$, which simplifies to $x^3-11x^2+30x-24=0$. Since $x=2$ is a root, the remaining quadratic is $x^2-9x+12=0$, so the other roots are $\frac{9\pm\sqrt{33}}{2}$. The cut length must satisfy $0<x<5$, so $\frac{9+\sqrt{33}}{2}\approx 7.37$ is invalid. The valid values are $2$ and $\frac{9-\sqrt{33}}{2}$.
17. How many real roots does $(x^2-1)^2(x^2+4)(x+3)^3=0$ have, counting multiplicities?
    Answer: 7 The factor $(x^2-1)^2$ gives roots $1$ and $-1$, each with multiplicity $2$, for $4$ real roots counted with multiplicity. The factor $x^2+4$ gives no real roots. The factor $(x+3)^3$ gives $-3$ with multiplicity $3$. Total: $4+3=7$. The choice $3$ counts distinct real roots; $4$ misses $-3$; $6$ is a degree-miscount.
18. After writing the volume equation and simplifying, a student gets $w^3 + w^2 - 6w - 84 = 0$. Between which two consecutive integers is the positive solution $w$?
    Answer: 4 and 5 When you see a volume equation, start by writing the product of the actual dimensions. Here the width is $w$, the height is $w-2$, and the length is $w+3$, so the volume equation is $$w(w-2)(w+3)=84.$$ Expanding gives $w^3+w^2-6w-84=0$, which matches the student's equation. To locate the positive solution, test consecutive integer boundaries. At $w=4$, the volume is $4(2)(7)=56$, still below 84. At $w=5$, the volume is $5(3)(8)=120$, already above 84. Since the volume grows as $w$ increases for positive $w$, the crossing must happen between 4 and 5. The interval "3 and 4" cannot be correct because even at $w=4$ the volume is too small. "5 and 6" and "6 and 7" are both past the crossing: at $w=5$ the volume already exceeds 84, and it continues increasing, so the root cannot lie there. A common trap is comparing 84 to $w^3$ alone, since $\sqrt[3]{84}\approx 4.4$; but the adjusted dimensions shift the true root. On exam day, when asked for an interval containing a polynomial root, evaluate the equation at the integer endpoints and look for a sign change. That will quickly point you to the correct consecutive integers.
19. One solution of $2x^3 - 4x^2 - 10x + 12 = 0$ is $x=1$. What is the product of the other two solutions?
    Answer: $-6$ Whenever a cubic equation gives you one root, you can use Vieta's formulas to avoid factoring everything by hand. For a cubic $ax^3+bx^2+cx+d=0$, the product of all three roots is $-\frac{d}{a}$. Here that product is $-\frac{12}{2}=-6$. Since one root is $x=1$, the product of the other two roots must be $-6 \div 1 = -6$. You can confirm by synthetic division: dividing $2x^3-4x^2-10x+12$ by $x-1$ gives $2x^2-2x-12$, which factors as $2(x-3)(x+2)$. So the other roots are $3$ and $-2$, and their product is $-6$. The choice $-2$ is just one of the other roots, not the product. The choice $3$ is the other root alone. The choice $6$ is the common sign mistake: it ignores the negative sign in Vieta's product formula or forgets that the leading coefficient affects the product. So when one root is given, remember: total product of all roots divided by the known root gives you the product of the remaining roots. That turns this into a one-step calculation.
20. Which statement correctly describes the real solutions of $(x-1)^2(x+2)^3(x^2+1)=0$?
    Answer: There are 2 distinct real solutions and 5 real solutions counting multiplicity. When you see a factored polynomial equation like this, your first move is to identify the roots from each factor and then check multiplicities. Also remember that factors like $x^2+1$ have no real roots, since $x^2=-1$ has no real solution. Here, the factor $(x-1)^2$ gives the real root $x=1$, and the exponent $2$ means it contributes $2$ to the real-solution count with multiplicity. The factor $(x+2)^3$ gives the real root $x=-2$, contributing $3$ with multiplicity. So there are $2$ distinct real solutions: $x=1$ and $x=-2$. Counting multiplicity, the total is $2+3=5$ real solutions. The statement claiming "5 distinct real solutions and 5 real solutions counting multiplicity" is wrong because there are only two distinct real roots; the $5$ is the multiplicity count, not the number of distinct roots. The statement claiming "2 distinct real solutions and 2 real solutions counting multiplicity" misses the multiplicities entirely—it counts each distinct root only once. The statement claiming "7 distinct real solutions and 7 real solutions counting multiplicity" confuses the total degree of the polynomial with real solutions; the $x^2+1$ factor contributes nonreal solutions $\pm i$, which do not count as real solutions. Strategy: when counting real solutions, first list the real roots from each factor, then add their multiplicities. Ignore nonreal factors like $x^2+1$ for real-solution counts.

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