Solving Polynomial Equations - Accuplacer Aaf

For the polynomial equation $ax^2 - bx - c = 0$ , the solutions are $x=2$ and $x=-1/3$ . Which of the following could be the factored form of the polynomial $ax^2 - bx - c$ ?
Answer: $(x-2)(3x+1)$ If the solutions (roots) of a polynomial equation are $r_1$ and $r_2$ , then the factored form of the polynomial is $a(x-r_1)(x-r_2)$ .\ Given the roots $r_1 = 2$ and $r_2 = -1/3$ , the factors are $(x-2)$ and $(x - (-1/3))$ , which simplifies to $(x+1/3)$ .\ So the factored form is $a(x-2)(x+1/3)$ . To eliminate the fraction and match the answer choices, we can choose a value for $a$ , such as $a=3$ .\ $3(x-2)(x+1/3) = (x-2) \cdot 3(x+1/3) = (x-2)(3x+1)\$ \ This corresponds to the polynomial $3x^2 - 5x - 2$ , which has the form $ax^2 - bx - c$ (with a=3, b=5, c=2). The factored form is $(x-2)(3x+1)$ .
How many distinct real solutions does the equation $x^3 + 27 = 0$ have?
Answer: 1 The equation is a sum of cubes, which factors as $a^3+b^3 = (a+b)(a^2-ab+b^2)$ .\ $$x^3$ + $3^3$ = (x+3)($x^2$ - 3x + 9) = 0\$ \ The first factor, $x+3=0$ , gives one real solution, $x=-3$ .\ For the second factor, $x^2 - 3x + 9 = 0$ , we check the discriminant, $b^2-4ac$ :\ $$(-3)^2$ - 4(1)(9) = 9 - 36 = -27\$ \ Since the discriminant is negative, this quadratic factor has no real solutions (it has two complex solutions). Therefore, the equation has only one distinct real solution.
What is the complete solution set for the equation $(x-5)(x+2) = x-5$ ?
Answer: \{-1, 5\} To solve, first set the equation to zero. Do not divide by $(x-5)$ as this would eliminate a solution.\ $(x-5)(x+2) - (x-5) = 0\$ \ Factor out the common term $(x-5)$ :\ $(x-5)((x+2) - 1) = 0\$ \ Simplify the expression in the second parenthesis:\ $(x-5)(x+1) = 0\$ \ Set each factor to zero: $x-5=0$ gives $x=5$ , and $x+1=0$ gives $x=-1$ .\ The complete solution set is $\{-1, 5\}$ .
What is the smallest integer solution to the equation $x^3 - 2x^2 = 9x - 18$ ?
Answer: -3 First, set the equation to zero.\ $$x^3$ - $2x^2$ - 9x + 18 = 0\$ \ Now, use factoring by grouping.\ $($x^3$ - $2x^2$) - (9x - 18) = 0\$ \ $$x^2$(x - 2) - 9(x - 2) = 0\$ \ $($x^2$ - 9)(x - 2) = 0\$ \ $(x-3)(x+3)(x-2) = 0\$ \ The solutions are $x=3$ , $x=-3$ , and $x=2$ . All are integers. The smallest of these is -3.
The equation $6x^3 - 19x^2 + 16x - 4 = 0$ has a rational root $x = \frac{1}{2}$ . What are all solutions to this equation?
Answer: $x = \frac{1}{2}, x = 2, x = \frac{2}{3}$ Since $x = \frac{1}{2}$ is a root, $(2x - 1)$ is a factor. Using synthetic or polynomial division: $6x^3 - 19x^2 + 16x - 4 = (2x - 1)(3x^2 - 8x + 4) = (2x - 1)(3x - 2)(x - 2)$ . Setting each factor to zero: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ , $3x - 2 = 0 \Rightarrow x = \frac{2}{3}$ , $x - 2 = 0 \Rightarrow x = 2$ . Choices B, C, and D contain incorrect factorizations of the quadratic factor.
If $x > 0$ , what is the solution to $x^3 + 3x^2 = 10x$ ?
Answer: 2 First, set the equation to zero by subtracting $10x$ from both sides.\ $$x^3$ + $3x^2$ - 10x = 0\$ \ Next, factor out the greatest common factor, $x$ :\ $x($x^2$ + 3x - 10) = 0\$ \ Factor the quadratic trinomial:\ $x(x+5)(x-2) = 0\$ \ The solutions are $x=0$ , $x=-5$ , and $x=2$ .\ The question specifies that $x > 0$ , so the only valid solution is $x=2$ .
If $2x^3 - 8x^2 - 18x + 72 = 0$ , and one solution is $x = 6$ , what is the sum of all solutions to this equation?
Answer: $4$ Since $x = 6$ is a solution, $(x - 6)$ is a factor. Using polynomial division: $2x^3 - 8x^2 - 18x + 72 = 2(x - 6)(x^2 + x - 6) = 2(x - 6)(x + 3)(x - 2)$ . The solutions are $x = 6, -3, 2$ , so their sum is $6 + (-3) + 2 = 4$ . Choice B uses only the given solution. Choice C miscalculates the sum. Choice D adds the absolute values.
The volume of a rectangular prism is represented by the polynomial $V(x) = x^3 + 5x^2 - 6x$ , where $x$ is a positive dimension. For which value of $x$ is the volume equal to zero?
Answer: 1 Set the volume polynomial equal to zero and solve for $x$ .\ $$x^3$ + $5x^2$ - 6x = 0\$ \ Factor out the greatest common factor, $x$ :\ $x($x^2$ + 5x - 6) = 0\$ \ Factor the quadratic trinomial:\ $x(x+6)(x-1) = 0\$ \ The solutions are $x=0$ , $x=-6$ , and $x=1$ . Since the problem states that $x$ is a positive dimension, the only valid answer is $x=1$ .
What is the sum of the real solutions for the equation $4x^4 - 9x^2 = 0$ ?
Answer: 0 First, factor out the greatest common factor, $x^2$ .\ $$x^2$($4x^2$ - 9) = 0\$ \ The second factor is a difference of squares, $(2x)^2 - 3^2$ .\ $$x^2$(2x - 3)(2x + 3) = 0\$ \ Set each factor to zero to find the solutions: $x^2=0 \implies x=0$ (a repeated root), $2x-3=0 \implies x=3/2$ , and $2x+3=0 \implies x=-3/2$ .\ The distinct real solutions are $0, 3/2, -3/2$ .\ The sum of the solutions is $0 + 3/2 + (-3/2) = 0$ .
The equation $x^3 - 4x^2 + 4x = 0$ has a real solution with a multiplicity of 2. What is the value of this solution?
Answer: 2 First, factor the polynomial. The greatest common factor is $x$ .\ $x($x^2$ - 4x + 4) = 0\$ \ The quadratic factor is a perfect square trinomial.\ $$x(x-2)^2$ = 0\$ \ The solutions are found by setting each factor to zero. $x=0$ is one solution. $(x-2)^2=0$ gives $x-2=0$ , so $x=2$ . Because the factor $(x-2)$ is squared, the root $x=2$ has a multiplicity of 2. Thus, 2 is the repeated solution.
What is the largest solution to the equation $x(x+1) = 12$ ?
Answer: 3 To solve the equation, first set it equal to zero.\ $x(x+1) = 12\$ \ $$x^2$ + x = 12\$ \ $$x^2$ + x - 12 = 0\$ \ Now, factor the quadratic trinomial.\ $(x+4)(x-3) = 0\$ \ The solutions are $x = -4$ and $x = 3$ . The largest of these solutions is 3.
If $f(x) = x^4 - 10x^2 + 9$ and $f(x) = 0$ , what is the product of all positive solutions?
Answer: $3$ When you encounter a polynomial equation set equal to zero, look for patterns that allow you to factor efficiently. This quartic equation has a special structure that makes it solvable. Notice that $f(x) = x^4 - 10x^2 + 9$ contains only even powers of $x$ . This suggests using substitution: let $u = x^2$ , so the equation becomes $u^2 - 10u + 9 = 0$ . This quadratic factors as $(u - 1)(u - 9) = 0$ , giving us $u = 1$ or $u = 9$ . Substituting back: when $x^2 = 1$ , we get $x = \pm 1$ , and when $x^2 = 9$ , we get $x = \pm 3$ . So the complete solution set is $x = -3, -1, 1, 3$ . The question asks for the product of all positive solutions. The positive solutions are $x = 1$ and $x = 3$ , so their product is $1 \times 3 = 3$ . Answer choice A ( $6$ ) represents the sum of the positive solutions ( $1 + 3 = 6$ ) rather than their product. Answer choice B ( $9$ ) is one of the intermediate values from our substitution ( $u = 9$ ), but not the final answer. Answer choice C ( $1$ ) is one of the positive solutions individually, not their product. The correct answer is D. Strategy tip: When you see polynomials with only even powers, try the substitution $u = x^2$ to reduce the degree by half. Also, always read carefully whether the question asks for sum, product, or individual solutions.
The equation $x^3 - 8 = 0$ has one real solution and two non-real complex solutions. What is the product of the two non-real solutions?
Answer: 4 Factor the difference of cubes: $x^3 - 2^3 = (x-2)(x^2 + 2x + 4) = 0$ .\ The real solution comes from $x-2=0$ , which is $x=2$ .\ The two non-real solutions come from the quadratic factor $x^2 + 2x + 4 = 0$ .\ For a quadratic equation $ax^2+bx+c=0$ , the product of the roots is given by $c/a$ .\ In this case, $a=1$ and $c=4$ . So, the product of the two non-real roots is $4/1 = 4$ .
Given that $x=3$ is one solution to the equation $x^3 - x^2 - 17x + 33 = 0$ , what is the sum of the other two solutions?
Answer: -2 If $x=3$ is a solution, then $(x-3)$ is a factor of the polynomial. We can use polynomial division or Vieta's formulas. Using Vieta's formulas, for a cubic equation $ax^3+bx^2+cx+d=0$ , the sum of the roots $r_1+r_2+r_3$ is $-b/a$ .\ In this equation, $a=1$ and $b=-1$ , so the sum of all three roots is $-(-1)/1 = 1$ .\ Let the roots be $r_1, r_2, r_3$ . We are given $r_1 = 3$ .\ So, $3 + r_2 + r_3 = 1$ .\ Subtracting 3 from both sides gives $r_2 + r_3 = 1 - 3 = -2$ . The sum of the other two solutions is -2.
The equation $8x^3 - 27 = 0$ has how many real solutions, and what is their sum?
Answer: One real solution; sum is $\frac{3}{2}$ When you encounter a cubic equation like this one, start by isolating the cubic term to see if you can solve it directly without factoring. Starting with $8x^3 - 27 = 0$ , add 27 to both sides: $8x^3 = 27$ . Then divide by 8: $x^3 = \frac{27}{8}$ . Taking the cube root of both sides gives $x = \sqrt[3]{\frac{27}{8}} = \frac{\sqrt[3]{27}}{\sqrt[3]{8}} = \frac{3}{2}$ . Unlike square roots, cube roots always produce exactly one real solution for any real number. Since $\left(\frac{3}{2}\right)^3 = \frac{27}{8}$ , we can verify this is correct. Therefore, there is one real solution, and its sum (since there's only one) is $\frac{3}{2}$ . Choice A incorrectly assumes two real solutions, likely confusing this with a quadratic equation or misremembering that even-degree polynomials can have multiple real roots. Choice B suggests three real solutions with sum zero, which might come from incorrectly factoring this as a difference of cubes and assuming all roots are real (complex cube roots do sum to zero, but they're not real). Choice C has the right number of solutions but the wrong value, possibly from calculation errors like mixing up the numerator and denominator. Remember that odd-degree polynomial equations always have at least one real solution, and simple cubic equations of the form $ax^3 = b$ have exactly one real solution found by taking the cube root.
The x-intercepts of the graph of a function $f(x)$ are the real solutions to $f(x)=0$ . How many distinct x-intercepts does the graph of $f(x) = x^4 + 6x^3 + 9x^2$ have?
Answer: 2 To find the x-intercepts, set $f(x) = 0$ and solve for $x$ .\ $$x^4$ + $6x^3$ + $9x^2$ = 0\$ \ Factor out the greatest common factor, $x^2$ :\ $$x^2$($x^2$ + 6x + 9) = 0\$ \ The expression in the parentheses is a perfect square trinomial.\ $$x^2$$(x+3)^2$ = 0\$ \ The solutions are found by setting the factors to zero: $x^2 = 0 \implies x=0$ and $(x+3)^2 = 0 \implies x=-3$ .\ Even though these roots have multiplicity 2, they correspond to only two distinct points on the x-axis. Thus, there are 2 distinct x-intercepts.
Which of the following is the complete solution set for the equation $5x^3 - 20x = 0$ ?
Answer: \{-2, 0, 2\} First, factor out the greatest common factor, $5x$ .\ $5x($x^2$ - 4) = 0\$ \ Next, factor the difference of squares, $x^2 - 4$ .\ $5x(x - 2)(x + 2) = 0\$ \ Using the Zero Product Property, set each factor equal to zero: $5x=0$ , $x-2=0$ , and $x+2=0$ .\ This gives the solutions $x=0$ , $x=2$ , and $x=-2$ . The complete solution set is $\{-2, 0, 2\}$ .
The polynomial $P(x) = x^4 - 5x^3 - 2x^2 + 24x$ can be completely factored. How many distinct real solutions does $P(x) = 0$ have?
Answer: Four distinct real solutions First factor out $x$ : $P(x) = x(x^3 - 5x^2 - 2x + 24)$ . For the cubic, try rational roots. Testing $x = 4$ : $64 - 80 - 8 + 24 = 0$ . So $(x - 4)$ is a factor. Dividing gives $x^3 - 5x^2 - 2x + 24 = (x - 4)(x^2 - x - 6) = (x - 4)(x - 3)(x + 2)$ . Therefore $P(x) = x(x - 4)(x - 3)(x + 2)$ , giving solutions $x = 0, 4, 3, -2$ (four distinct values). Other choices undercount the solutions.
What is the sum of the positive real solutions to the equation $x^4 - 5x^2 + 4 = 0$ ?
Answer: 3 This equation is in quadratic form. Let $u = x^2$ . The equation becomes:\ $$u^2$ - 5u + 4 = 0\$ \ Factor the quadratic:\ $(u-4)(u-1) = 0\$ \ This gives $u=4$ or $u=1$ . Substitute back $x^2$ for $u$ :\ $$x^2$ = 4 \quad \text{or} \quad $x^2$ = 1\$ \ Solving for $x$ gives $x = \pm 2$ and $x = \pm 1$ . The positive real solutions are 1 and 2. Their sum is $1 + 2 = 3$ .
For what values of $k$ does the equation $x^3 - 7x^2 + kx - 12 = 0$ have $x = 3$ as a solution?
Answer: $k = 16$ only If $x = 3$ is a solution, then substituting gives: $3^3 - 7(3^2) + k(3) - 12 = 0$ , so $27 - 63 + 3k - 12 = 0$ , which simplifies to $3k - 48 = 0$ , giving $k = 16$ . Choice A comes from incorrectly calculating $7 \times 9 = 56$ instead of $63$ . Choice C suggests multiple values exist. Choice D confuses the constant term with the parameter.

Start this quiz