Factoring to Solve Equations - Accuplacer Aaf

If $2x^2 + 8x = x^2 + 3x + 10$ , what is the sum of all solutions?
Answer: $-5$ Rearrange to standard form: $2x^2 + 8x - x^2 - 3x - 10 = 0$ , which gives $x^2 + 5x - 10 = 0$ . Using the sum formula for quadratic roots, the sum is $-\frac{b}{a} = -\frac{5}{1} = -5$ . Choice A uses the positive coefficient. Choice C comes from incorrectly combining like terms. Choice D assumes the constant term affects the sum.
What is the complete set of real solutions for the equation $8x^3 + 27 = 0$ ?
Answer: $\{-\frac{3}{2}\}$ This equation is a sum of cubes: $(2x)^3 + 3^3 = 0$ . The sum of cubes formula is $a^3+b^3 = (a+b)(a^2-ab+b^2)$ . Applying this gives $(2x+3)(4x^2 - 6x + 9) = 0$ . The first factor, $2x+3=0$ , yields the real solution $x = -\frac{3}{2}$ . The second factor, $4x^2 - 6x + 9 = 0$ , has a discriminant of $b^2-4ac = (-6)^2 - 4(4)(9) = 36 - 144 = -108$ . Since the discriminant is negative, there are no other real solutions.
The equation $4x^2 + kx + 9 = 0$ has exactly one real solution. If $k > 0$ , what is the value of the solution?
Answer: $-\frac{3}{2}$ For a quadratic equation to have exactly one real solution, it must be a perfect square trinomial. The form is $(ax+b)^2 = a^2x^2 + 2abx + b^2$ . Here, $a^2=4$ so $a=2$ , and $b^2=9$ so $b=3$ . The middle term $kx$ must be $2abx = 2(2)(3)x = 12x$ . So, $k=12$ . The equation is $4x^2 + 12x + 9 = 0$ , which factors to $(2x+3)^2=0$ . Solving for $x$ gives $2x+3=0$ , so $x = -\frac{3}{2}$ .
The solutions to the equation $x^2 - 8x + 12 = 0$ are $x=2$ and $x=6$ . Which of the following equations has solutions that are each 3 greater than the solutions of the original equation?
Answer: $x^2 - 14x + 45 = 0$ The original solutions are 2 and 6. The new solutions must be 3 greater than these, so they are $2+3=5$ and $6+3=9$ . To find the equation with these solutions, use the factors $(x-5)$ and $(x-9)$ . The equation is $(x-5)(x-9) = 0$ . Expanding this gives $x^2 - 9x - 5x + 45 = 0$ , which simplifies to $x^2 - 14x + 45 = 0$ .
What is the sum of all solutions to the equation $x - 8\sqrt{x} + 15 = 0$ ?
Answer: $34$ This equation is quadratic in form. Let $u = \sqrt{x}$ , so $u^2 = x$ . The equation becomes $u^2 - 8u + 15 = 0$ . Factoring gives $(u-3)(u-5)=0$ , so $u=3$ or $u=5$ . Substitute back: $\sqrt{x}=3$ or $\sqrt{x}=5$ . Squaring both sides of each equation gives $x=9$ and $x=25$ . Both solutions are valid. The sum of the solutions is $9 + 25 = 34$ .
If $x^3 + 2x^2 - 9x - 18 = 0$ , what is the product of the solutions?
Answer: $18$ This polynomial can be factored by grouping: $(x^3 + 2x^2) - (9x + 18) = 0$ . Factoring out common terms from each group gives $x^2(x + 2) - 9(x + 2) = 0$ . This can be written as $(x^2 - 9)(x + 2) = 0$ . Factoring the difference of squares gives $(x - 3)(x + 3)(x + 2) = 0$ . The solutions are $x = 3$ , $x = -3$ , and $x = -2$ . The product is $(3)(-3)(-2) = 18$ .
If $(x + 3)(2x - 1) = x + 3$ , what are all possible values of $x$ ?
Answer: $x = -3$ or $x = 1$ When you encounter an equation where one side is a factor of the other, like $(x + 3)(2x - 1) = x + 3$ , you're dealing with a factoring situation that requires careful algebraic manipulation. To solve this equation, start by moving everything to one side: $(x + 3)(2x - 1) - (x + 3) = 0$ . Now you can factor out the common term $(x + 3)$ : $(x + 3)[(2x - 1) - 1] = 0$ , which simplifies to $(x + 3)(2x - 2) = 0$ , or $(x + 3) \cdot 2(x - 1) = 0$ . Using the zero product property, either $x + 3 = 0$ or $x - 1 = 0$ . This gives you $x = -3$ or $x = 1$ . You can verify: when $x = -3$ , both sides equal $0$ ; when $x = 1$ , both sides equal $4$ . Looking at the wrong answers: Choice A ( $x = -3$ only) misses the second solution by failing to apply the zero product property completely. Choice B ( $x = 1$ only) makes the same error in reverse. Choice C ( $x = -3$ or $x = \frac{1}{2}$ ) likely results from an arithmetic error when solving $2x - 1 = 1$ , incorrectly getting $2x = \frac{1}{2}$ instead of $2x = 2$ . The correct answer is D. Remember: when you have a product equal to a single term, factor out common expressions first, then use the zero product property. Always check both solutions by substituting back into the original equation.
The solutions to $5x^2 = 20x$ are $x=0$ and $x=k$ . What is the value of $k$ ?
Answer: $4$ To solve for all solutions, first set the equation to zero: $5x^2 - 20x = 0$ . Do not divide by $x$ , as this will lose a solution. Instead, factor out the greatest common factor, $5x$ : $5x(x - 4) = 0$ . The solutions are found by setting each factor to zero: $5x = 0$ gives $x=0$ , and $x - 4 = 0$ gives $x=4$ . Therefore, $k=4$ .
What is the smallest solution to the equation $30 + 5x - 5x^2 = 0$ ?
Answer: $-2$ First, rearrange the equation into standard form and factor out the greatest common factor, -5: $-5x^2 + 5x + 30 = 0$ becomes $-5(x^2 - x - 6) = 0$ . Divide by -5 to get $x^2 - x - 6 = 0$ . Factoring the trinomial gives $(x-3)(x+2) = 0$ . The solutions are $x=3$ and $x=-2$ . The smallest of these solutions is $-2$ .
If $3x^2 - 12x = 0$ and $x \neq 0$ , what is the value of $x + 2$ ?
Answer: $6$ Factor: $3x(x - 4) = 0$ , giving solutions $x = 0$ and $x = 4$ . Since $x \neq 0$ , we have $x = 4$ . Therefore $x + 2 = 4 + 2 = 6$ . Choice A comes from using $x = 0$ despite the constraint. Choice B gives the value of $x$ rather than $x + 2$ . Choice D adds incorrectly or uses wrong solution.
For which value of $x$ does $(x - 2)^2 - 9 = 0$ ?
Answer: $x = 5$ or $x = -1$ Recognize this as a difference of squares: $(x - 2)^2 - 9 = (x - 2)^2 - 3^2 = [(x - 2) - 3][(x - 2) + 3] = (x - 5)(x + 1) = 0$ . This gives $x = 5$ or $x = -1$ . Choices A and B identify only one solution each. Choice D comes from incorrectly expanding or factoring the expression.
If $x^2 + 6x + 9 = 25$ , how many distinct solutions does the equation have?
Answer: Exactly two solutions Rearrange: $x^2 + 6x + 9 - 25 = 0$ , so $x^2 + 6x - 16 = 0$ . Factor: $(x + 8)(x - 2) = 0$ , giving $x = -8$ and $x = 2$ . There are exactly two distinct solutions. Choice A occurs if students think the discriminant is negative. Choice B might result from recognizing the perfect square $(x + 3)^2$ but not properly handling the constant. Choice D is not possible for a quadratic equation.
The equation $4x^2 - 4x - 8 = 0$ can be solved by factoring. What is the positive solution?
Answer: $x = 2$ First factor out the GCF: $4(x^2 - x - 2) = 0$ , so $x^2 - x - 2 = 0$ . Factor the quadratic: $(x - 2)(x + 1) = 0$ . The solutions are $x = 2$ and $x = -1$ . The positive solution is $x = 2$ . Choice A comes from incorrectly factoring or solving. Choice C uses the leading coefficient. Choice D uses the constant term from the original equation.
One of the factors of $x^4 - 81$ is $x^2 + 9$ . What is the sum of the real solutions to the equation $x^4 - 81 = 0$ ?
Answer: $0$ The expression $x^4 - 81$ is a difference of squares, $(x^2)^2 - 9^2$ , which factors to $(x^2 - 9)(x^2 + 9) = 0$ . The factor $x^2 + 9 = 0$ has no real solutions. The factor $x^2 - 9 = 0$ is another difference of squares, $(x-3)(x+3)=0$ . This gives the real solutions $x=3$ and $x=-3$ . The sum of these solutions is $3 + (-3) = 0$ .
How many distinct real solutions does the equation $16x^4 - 1 = 0$ have?
Answer: $2$ The equation is a difference of squares: $(4x^2)^2 - 1^2 = 0$ . It factors into $(4x^2 - 1)(4x^2 + 1) = 0$ . The second factor, $4x^2 + 1 = 0$ , has no real solutions because $x^2$ cannot be negative. The first factor, $4x^2 - 1 = 0$ , is also a difference of squares, factoring into $(2x - 1)(2x + 1) = 0$ . This yields two distinct real solutions: $x = \frac{1}{2}$ and $x = -\frac{1}{2}$ .
Which of the following is a solution to the equation $(2x - 1)^2 = 49$ ?
Answer: $-3$ The equation can be solved by setting it to zero and factoring as a difference of squares: $(2x - 1)^2 - 49 = 0$ , which is $(2x - 1 - 7)(2x - 1 + 7) = 0$ . This simplifies to $(2x - 8)(2x + 6) = 0$ . The solutions are from $2x - 8 = 0$ (which gives $x = 4$ ) and $2x + 6 = 0$ (which gives $x = -3$ ). Of the choices provided, $-3$ is a solution.
If $x^2 - 3x = 10$ , what is the sum of the possible values of $x$ ?
Answer: $3$ To solve the equation, first set it to zero: $x^2 - 3x - 10 = 0$ . Factoring the quadratic gives $(x - 5)(x + 2) = 0$ . The solutions are $x = 5$ and $x = -2$ . The sum of these values is $5 + (-2) = 3$ .
If $x^2 - 5x = 6$ , what is the product of all possible values of $x$ ?
Answer: $-6$ First, rearrange to standard form: $x^2 - 5x - 6 = 0$ . Factor: $(x - 6)(x + 1) = 0$ . The solutions are $x = 6$ and $x = -1$ . The product is $6 \times (-1) = -6$ . Choice B assumes the constant term is the product. Choice C uses the coefficient of the linear term. Choice D multiplies the solutions incorrectly as positive values.
The length of a rectangular garden is 1 foot less than twice its width. Its area is 120 square feet. What is its width in feet?
Answer: $8$ Let $w$ be the width. The length $l$ is $2w-1$ . The area is $w(2w-1) = 120$ . This gives the equation $2w^2 - w - 120 = 0$ . Factoring this quadratic (by grouping or trial and error) gives $(2w+15)(w-8) = 0$ . The possible solutions are $w = -\frac{15}{2}$ or $w=8$ . Since width must be positive, the width is 8 feet.
What is the solution set for the equation $x - 6 = \frac{16}{x}$ ?
Answer: $\{-2, 8\}$ Assuming $x \neq 0$ , multiply both sides by $x$ to clear the denominator: $x(x-6) = 16$ , which simplifies to $x^2 - 6x = 16$ . Set the equation to zero: $x^2 - 6x - 16 = 0$ . Factoring the quadratic gives $(x-8)(x+2) = 0$ . The solutions are $x = 8$ and $x = -2$ .

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