Factoring Trinomials - Accuplacer Aaf

1. If $16x^2 - kx + 9$ is a perfect square trinomial, which of the following could be the value of $k$?
   Answer: 24 A perfect square trinomial has the form $a^2 - 2ab + b^2 = (a-b)^2$ or $a^2 + 2ab + b^2 = (a+b)^2$. In $16x^2 - kx + 9$, we have $a^2 = 16x^2$, so $a=4x$, and $b^2 = 9$, so $b=3$. The middle term, $-kx$, must be equal to $-2ab$. Substituting the values for $a$ and $b$, we get $-2(4x)(3) = -24x$. Therefore, $-kx = -24x$, which means $k=24$.
2. Which of the following is equivalent to the expression $\frac{2x^2 - x - 15}{x^2 - 9}$ for all $x \neq \pm3$?
   Answer: $\frac{2x+5}{x+3}$ To simplify the expression, factor the numerator and the denominator. The numerator, $2x^2 - x - 15$, factors into $(2x+5)(x-3)$. The denominator, $x^2 - 9$, is a difference of squares and factors into $(x-3)(x+3)$. The expression becomes $\frac{(2x+5)(x-3)}{(x-3)(x+3)}$. The $(x-3)$ terms cancel, leaving $\frac{2x+5}{x+3}$.
3. The expression $3(x+2)^2 - 5(x+2) - 2$ can be factored into the form $(ax+b)(cx+d)$. What is the value of $a+c$?
   Answer: 4 Let $u = x+2$. The expression becomes $3u^2 - 5u - 2$. We can factor this trinomial. We need two numbers that multiply to $3 \times -2 = -6$ and add to -5. These numbers are -6 and 1. So, $3u^2 - 6u + u - 2 = 3u(u-2) + 1(u-2) = (3u+1)(u-2)$. Now substitute back $u=x+2$: $(3(x+2)+1)((x+2)-2) = (3x+6+1)(x) = (3x+7)(x)$. The factors are $x$ and $3x+7$. We can write $x$ as $1x+0$. So the factored form is $(3x+7)(1x+0)$. We have $a=3, c=1$ (or vice versa). The sum $a+c = 3+1 = 4$.
4. Which of the following is a factor of the expression $x^2 - 3xy - 10y^2$?
   Answer: $x - 5y$ To factor the trinomial $x^2 - 3xy - 10y^2$, we look for two numbers that multiply to -10 and add to -3. These numbers are -5 and 2. The factorization is $(x - 5y)(x + 2y)$. Therefore, $x - 5y$ is one of the factors.
5. Which of the following is the correct factorization of $10x^2 - 11x - 6$?
   Answer: $(5x+2)(2x-3)$ To factor $10x^2 - 11x - 6$, we look for two numbers that multiply to $10 \times -6 = -60$ and add to -11. These numbers are -15 and 4. We can rewrite the middle term: $10x^2 - 15x + 4x - 6$. Factor by grouping: $5x(2x - 3) + 2(2x - 3) = (5x+2)(2x-3)$.
6. If $2x^2 - 8x + k$ can be factored as $(2x - a)(x - b)$ where $a$ and $b$ are positive integers, what is the value of $k$?
   Answer: 6 Expanding $(2x - a)(x - b) = 2x^2 - 2bx - ax + ab = 2x^2 - (2b + a)x + ab$. Comparing with $2x^2 - 8x + k$, we need $2b + a = 8$ and $k = ab$. Since $a$ and $b$ are positive integers, possible pairs are: $(a,b) = (2,3), (4,2), (6,1)$. Only $(2,3)$ gives $2(3) + 2 = 8$, so $k = 2 \cdot 3 = 6$.
7. For which value of $n$ is $x^2 + nx + 36$ factorable as $(x + a)(x + b)$ where both $a$ and $b$ are positive integers?
   Answer: $n = 37$ If $x^2 + nx + 36 = (x + a)(x + b)$ where $a, b > 0$, then $ab = 36$ and $a + b = n$. The positive factor pairs of 36 are: $(1,36), (2,18), (3,12), (4,9), (6,6)$. Their sums are: $37, 20, 15, 13, 12$ respectively. Among the choices, only $n = 37$ works, corresponding to $(x + 1)(x + 36)$.
8. If $4x^2 + 12x + 9 = (px + q)^2$ for all values of $x$, what is $p + q$?
   Answer: 5 Since $4x^2 + 12x + 9$ is a perfect square trinomial, we recognize it as $(2x + 3)^2$. We can verify: $(2x + 3)^2 = 4x^2 + 12x + 9$. Comparing with $(px + q)^2$, we have $p = 2$ and $q = 3$, so $p + q = 5$.
9. If $ax^2 + bx + c$ factors as $(px + q)(rx + s)$, which of the following must be true?
   Answer: $a = pr$, $b = ps + qr$, and $c = qs$ When you see a question about factoring quadratic expressions, you're dealing with the relationship between expanded and factored forms. The key is understanding how the distributive property (FOIL) connects the coefficients. To find the correct relationships, expand $(px + q)(rx + s)$ using FOIL: - First terms: $px \cdot rx = prx^2$ - Outer terms: $px \cdot s = psx$ - Inner terms: $q \cdot rx = qrx$ - Last terms: $q \cdot s = qs$ This gives us: $prx^2 + psx + qrx + qs = prx^2 + (ps + qr)x + qs$ Comparing this to $ax^2 + bx + c$, we can match coefficients: - Coefficient of $x^2$: $a = pr$ - Coefficient of $x$: $b = ps + qr$ - Constant term: $c = qs$ Choice A is incomplete because it only gives two of the three necessary relationships and omits the middle term coefficient. Choice B incorrectly swaps the relationships, claiming $a = ps + qr$ (which should be $b$) and $c = pr$ (which should be $a$). Choice C suggests addition relationships ($a = p + r$ and $c = q + s$) that don't reflect how multiplication works in polynomial expansion. Choice D correctly identifies all three relationships: $a = pr$, $b = ps + qr$, and $c = qs$. Remember: when factoring quadratics, the leading coefficient comes from multiplying the $x$-coefficients in each factor, the constant comes from multiplying the constant terms, and the middle coefficient is the cross-sum of outer and inner products.
10. The expression $12x^2 + kx - 15$ can be written as $(3x + 5)(4x - 3)$. What is the value of $k$?
    Answer: $k = 11$ Expanding $(3x + 5)(4x - 3) = 12x^2 - 9x + 20x - 15 = 12x^2 + 11x - 15$. Comparing with $12x^2 + kx - 15$, we see that $k = 11$.
11. When the expression $4x^3 + 10x^2 - 6x$ is factored completely, which of the following is one of its factors?
    Answer: $2x-1$ First, factor out the greatest common factor (GCF), which is $2x$. This gives $2x(2x^2 + 5x - 3)$. Next, factor the trinomial $2x^2 + 5x - 3$. We look for two numbers that multiply to $2 \times -3 = -6$ and add to 5. These numbers are 6 and -1. This leads to the factorization $(2x-1)(x+3)$. The complete factorization is $2x(2x-1)(x+3)$. The factors are $2x$, $2x-1$, and $x+3$. Of the choices provided, $2x-1$ is a factor.
12. If $x+4$ is a factor of the trinomial $x^2 + kx - 24$, what is the other factor?
    Answer: $x-6$ If $x+4$ is a factor, then the factorization must be of the form $(x+4)(x+n)$. Expanding this gives $x^2 + (4+n)x + 4n$. Comparing this to $x^2 + kx - 24$, we can see that $4n = -24$, which means $n = -6$. Therefore, the other factor is $x-6$. The full factorization is $(x+4)(x-6) = x^2 - 2x - 24$, so $k = -2$.
13. A rectangular prism has a volume of $3x^3 + 17x^2 + 10x$. If the height is $x$, which of the following could represent the length and width of the prism's base?
    Answer: $3x+2$ and $x+5$ The volume of a rectangular prism is $V = \text{length} \times \text{width} \times \text{height}$. We are given $V = 3x^3 + 17x^2 + 10x$ and height $= x$. The area of the base is $V/h = (3x^3 + 17x^2 + 10x)/x = 3x^2 + 17x + 10$. To find the length and width, we must factor this trinomial. We need two numbers that multiply to $3 \times 10 = 30$ and add to 17. These numbers are 15 and 2. Rewriting the trinomial: $3x^2 + 15x + 2x + 10 = 3x(x+5) + 2(x+5) = (3x+2)(x+5)$. Thus, the length and width could be $3x+2$ and $x+5$.
14. The trinomial $x^2 + bx - 18$ can be factored into two binomials with integer coefficients. Which of the following is NOT a possible value for $b$?
    Answer: 12 For the trinomial to be factorable, $b$ must be the sum of a pair of integer factors of -18. The factor pairs of -18 are (1, -18), (-1, 18), (2, -9), (-2, 9), (3, -6), and (-3, 6). The possible sums (values for $b$) are -17, 17, -7, 7, -3, and 3. Of the given choices, -17, -7, and 3 are all possible values for $b$. The value 12 is not a possible sum of any integer factor pair of -18.
15. Three of the following trinomials can be factored over the integers. Which one CANNOT?
    Answer: $2x^2 + 3x - 4$ We test each option. A) $2x^2 + 7x + 3 = (2x+1)(x+3)$. B) $x^2 - 5x - 14 = (x-7)(x+2)$. C) $3x^2 - 5x + 2 = (3x-2)(x-1)$. D) For $2x^2 + 3x - 4$, we need two integers that multiply to $2 \times -4 = -8$ and sum to 3. The integer factor pairs of -8 are (1, -8), (-1, 8), (2, -4), and (-2, 4). None of these pairs sum to 3, so the trinomial cannot be factored over the integers.
16. The area of a rectangular garden is represented by the expression $3x^2 + 10x - 8$. If the width of the garden is represented by $x+4$, what expression represents the length?
    Answer: $3x-2$ The area of a rectangle is length times width. To find the length, we must factor the area expression $3x^2 + 10x - 8$. Since we know one factor is $x+4$, the other factor must be of the form $(3x+k)$ to get the $3x^2$ term. The product of the constant terms must be -8, so $4k = -8$, which gives $k = -2$. Thus, the other factor (the length) is $3x-2$. Checking the expansion: $(x+4)(3x-2) = 3x^2 - 2x + 12x - 8 = 3x^2 + 10x - 8$.
17. The trinomial $x^2 - 6x + 5$ and the trinomial $2x^2 - 7x - 15$ share a common binomial factor. What is that common factor?
    Answer: $x-5$ First, factor the simpler trinomial, $x^2 - 6x + 5$. This factors into $(x-1)(x-5)$. The common factor must be either $x-1$ or $x-5$. Now, factor the second trinomial, $2x^2 - 7x - 15$. This factors into $(2x+3)(x-5)$. Comparing the factors of both trinomials, the common factor is $x-5$.
18. For the trinomial $ax^2 + bx + c$, if $a=3$ and $c=-4$, which of the following values of $b$ allows the trinomial to be factored over the integers?
    Answer: -11 The trinomial is $3x^2 + bx - 4$. For it to be factorable, $b$ must be the sum of the inner and outer products of its binomial factors. The factors of $a=3$ are (1, 3). The factors of $c=-4$ are (1, -4), (-1, 4), (2, -2). We test the combinations: $(x+1)(3x-4) \implies b = -4+3 = -1$. $(x-4)(3x+1) \implies b = 1-12 = -11$. $(x-1)(3x+4) \implies b = 4-3 = 1$. $(x+4)(3x-1) \implies b = -1+12 = 11$. $(x+2)(3x-2) \implies b = -2+6 = 4$. $(x-2)(3x+2) \implies b = 2-6 = -4$. Of the choices provided, -11 is a possible value for $b$.
19. The expression $(5-x)(x+2)$ is the factored form of which of the following trinomials?
    Answer: $-x^2 + 3x + 10$ To find the trinomial, expand the factored expression using the FOIL method: $(5-x)(x+2) = (5)(x) + (5)(2) + (-x)(x) + (-x)(2) = 5x + 10 - x^2 - 2x$. Combining like terms gives $-x^2 + (5x - 2x) + 10 = -x^2 + 3x + 10$.
20. The trinomial $2x^2 + 9x - 5$ is factored into the form $(ax+b)(cx+d)$. If $a > 0$ and $c > 0$, what is the value of the sum of the four coefficients, $a+b+c+d$?
    Answer: 7 First, factor the trinomial $2x^2 + 9x - 5$. We need two numbers that multiply to $2 \times -5 = -10$ and add to 9. These numbers are 10 and -1. Rewriting the trinomial and factoring by grouping: $2x^2 + 10x - x - 5 = 2x(x+5) - 1(x+5) = (2x-1)(x+5)$. The factors are $(2x-1)$ and $(x+5)$. So, we can set $a=2, b=-1, c=1, d=5$. The sum of the coefficients is $a+b+c+d = 2 + (-1) + 1 + 5 = 7$.

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