Factoring Polynomials - Math 2

1. A polynomial $P(x) = ax^2 + bx + c$ has the property that $P(2) = 0$ and $P(-3) = 0$. If the leading coefficient is $-2$, what is the factored form of $P(x)$?
   Answer: $-2(x - 2)(x + 3)$ Since $P(2) = 0$ and $P(-3) = 0$, the factors are $(x-2)$ and $(x+3)$. With leading coefficient $-2$, we get $P(x) = -2(x-2)(x+3)$. Choice A switches the signs in the factors. Choice C has both factors with wrong signs. Choice D has both zeros wrong.
2. If $x^2 + kx + 36$ can be factored as $(x + a)(x + b)$ where $a$ and $b$ are positive integers, which of the following could be the value of $k$?
   Answer: $k = 15$ For $(x + a)(x + b) = x^2 + (a+b)x + ab$, we need $ab = 36$ and $a + b = k$ with $a, b$ positive integers. Factor pairs of 36: (1,36), (2,18), (3,12), (4,9), (6,6). The sums are 37, 20, 15, 13, 12. Among the choices, both $k = 15$ (from $a = 3, b = 12$) and $k = 20$ (from $a = 2, b = 18$) are possible, but only $k = 15$ matches choice B.
3. If $(x-3)(x+1)(x-5) = x^3 + ax^2 + bx + c$, then $a + b + c =$
   Answer: $15$ We can find $a + b + c$ by substituting $x = 1$ into both sides of the equation. Left side: $(1-3)(1+1)(1-5) = (-2)(2)(-4) = 16$. Right side: $1^3 + a(1)^2 + b(1) + c = 1 + a + b + c$. Therefore $1 + a + b + c = 16$, so $a + b + c = 15$. Alternatively, expanding directly: $(x-3)(x+1) = x^2 - 2x - 3$, then $(x^2 - 2x - 3)(x - 5) = x^3 - 5x^2 - 2x^2 + 10x - 3x + 15 = x^3 - 7x^2 + 7x + 15$. So $a = -7$, $b = 7$, $c = 15$, and $a + b + c = -7 + 7 + 15 = 15$.
4. Which of the following expressions is NOT factorable using integer coefficients?
   Answer: $x^4 + x^2 + 1$ Choice A: $9x^4 - 16y^4 = (3x^2)^2 - (4y^2)^2 = (3x^2 - 4y^2)(3x^2 + 4y^2)$ factors using integers. Choice B: $x^4 + 4x^2 + 4 = (x^2 + 2)^2$ is a perfect square. Choice C: Let $u = x^2$, then $u^2 - 5u + 6 = (u-2)(u-3) = (x^2-2)(x^2-3)$ factors with integer coefficients. Choice D: $x^4 + x^2 + 1$ cannot be factored with integer coefficients. Using $u = x^2$: $u^2 + u + 1$ has discriminant $1 - 4 = -3 < 0$, so it has no rational roots and cannot be factored over the integers.
5. When factoring $3x^3 - 12x^2 + 12x$, a student first factors out the greatest common factor and then factors the remaining expression. What is the complete factorization?
   Answer: $3x(x - 2)^2$ First factor out the GCF of $3x$: $3x(x^2 - 4x + 4)$. Then recognize that $x^2 - 4x + 4 = (x-2)^2$, giving $3x(x-2)^2$. Choice B stops after factoring the GCF. Choice C incorrectly factors the trinomial. Choice D has an incorrect GCF.
6. When factoring $2x^2 - 50$, what is the first step a student should take?
   Answer: Factor out the greatest common factor of 2 First factor out the GCF of 2: $2x^2 - 50 = 2(x^2 - 25)$. Then apply difference of squares to get $2(x-5)(x+5)$. Choice B skips the GCF step. Choice C is unnecessarily complex. Choice D applies trinomial factoring incorrectly to this binomial.
7. Which expression is equivalent to $4x^4 - 36x^2 + 81$?
   Answer: $(2x^2 - 9)^2$ This is a perfect square trinomial in terms of $x^2$. Let $u = x^2$, then we have $4u^2 - 36u + 81$. We need to check if this fits the pattern $a^2 - 2ab + b^2 = (a-b)^2$. Here $a = 2u = 2x^2$ and $b = 9$, so $2ab = 2(2x^2)(9) = 36x^2$. This gives us $(2x^2 - 9)^2$. Choice B expands to $4x^4 - 36x^2 + 81$ but is unnecessarily complex, choice C doesn't match when expanded, and choice D expands to $4x^4 - 8x^2 + 18x^2 - 72 = 4x^4 + 10x^2 - 72$.
8. A rectangular garden has area $6x^2 + 19x + 15$ square feet. If the length is $(2x + 3)$ feet, what is the width?
   Answer: $(3x + 5)$ feet When you see a rectangle problem involving polynomials, you're working with the fundamental relationship: Area = Length × Width. This means Width = Area ÷ Length, so you need to divide the area polynomial by the length polynomial. To find the width, divide $6x^2 + 19x + 15$ by $(2x + 3)$. You can use polynomial long division, but factoring the area expression is often faster. Look for two binomials that multiply to give $6x^2 + 19x + 15$. Since we know one factor is $(2x + 3)$, we need $(2x + 3) \times (\text{something}) = 6x^2 + 19x + 15$. The "something" must start with $3x$ (since $2x \times 3x = 6x^2$) and end with $+5$ (since $3 \times 5 = 15$). Testing $(3x + 5)$: $(2x + 3)(3x + 5) = 6x^2 + 10x + 9x + 15 = 6x^2 + 19x + 15$ ✓ Therefore, the width is $(3x + 5)$ feet, which is choice D. Choice A gives $(2x + 3)(4x + 5) = 8x^2 + 22x + 15$, which doesn't match our area. Choice B gives $(2x + 3)(3x - 5) = 6x^2 - x - 15$, with wrong middle and last terms. Choice C gives $(2x + 3)(6x + 5) = 12x^2 + 28x + 15$, producing the wrong leading coefficient and middle term. Remember: when working with polynomial area problems, always verify your answer by multiplying length times width to confirm you get back the original area expression.
9. If $x^2 - y^2 = (x-y)(x+y)$, then which expression represents the factorization of $(a+b)^2 - (c-d)^2$?
   Answer: $(a + b - c + d)(a + b + c - d)$ When you encounter problems involving the difference of squares formula, remember that the pattern $x^2 - y^2 = (x-y)(x+y)$ can be applied to any two expressions, not just simple variables. Here, you need to factor $(a+b)^2 - (c-d)^2$. This fits the difference of squares pattern where your first term is $(a+b)$ and your second term is $(c-d)$. Applying the formula: $(a+b)^2 - (c-d)^2 = [(a+b) - (c-d)][(a+b) + (c-d)]$. Now simplify each factor carefully. The first factor becomes: $(a+b) - (c-d) = a+b-c+d$ (remember that subtracting $(c-d)$ means you distribute the negative sign). The second factor becomes: $(a+b) + (c-d) = a+b+c-d$. This gives you $(a+b-c+d)(a+b+c-d)$, which matches answer choice D. Choice A incorrectly places the terms as $(a+b+c-d)(a+b-c+d)$, which reverses the order of the factors. Choice B gives $(a+b-c-d)(a+b+c+d)$, showing an error in sign distribution when subtracting $(c-d)$. Choice C has $(a-b+c-d)(a+b-c+d)$, which suggests treating this as $(a-b)^2$ instead of $(a+b)^2$. The key strategy here is to identify what plays the role of "x" and "y" in the difference of squares formula, then carefully distribute signs when subtracting. Always double-check your sign work when expanding expressions with parentheses.
10. The expression $8x^6 - 27y^3$ can be factored as a difference of cubes. What is the complete factorization?
    Answer: $(2x^2 - 3y)(4x^4 + 6x^2y + 9y^2)$ When you see an expression like $8x^6 - 27y^3$, recognize this as a difference of cubes pattern: $a^3 - b^3 = (a-b)(a^2 + ab + b^3)$. The key is identifying what's being cubed. First, rewrite the expression to see the cube structure clearly: $8x^6 - 27y^3 = (2x^2)^3 - (3y)^3$. Here, $a = 2x^2$ and $b = 3y$. Now apply the difference of cubes formula: - First factor: $(a-b) = (2x^2 - 3y)$ - Second factor: $a^2 + ab + b^2 = (2x^2)^2 + (2x^2)(3y) + (3y)^2 = 4x^4 + 6x^2y + 9y^2$ This gives us $(2x^2 - 3y)(4x^4 + 6x^2y + 9y^2)$, which is answer choice C. Answer choice A has the wrong sign in the second factor ($-6x^2y$ instead of $+6x^2y$). This represents the common error of using the difference of squares pattern instead of difference of cubes. Answer choice B incorrectly identifies the cubed terms as $2x^3$ and $3y$, leading to wrong exponents throughout the factorization. Answer choice D isn't actually factored at all—it just rewrites the original expression showing the cube structure but doesn't complete the factorization process. Remember: difference of cubes always has a PLUS sign in the trinomial factor ($a^2 + ab + b^2$), unlike difference of squares. Double-check your identification of the cubed terms before applying the formula.
11. Which of the following expressions cannot be factored over the integers?
    Answer: $x^2 + 6x + 10$ For $x^2 + 6x + 10$, we need integers that multiply to 10 and add to 6. The factor pairs of 10 are (1,10) and (2,5), with sums 11 and 7 respectively. Neither sum equals 6, so this cannot be factored over integers. The other expressions factor as: A: $(x+3)^2$, B: $(x+2)(x+4)$, D: $(x+1)(x+5)$.
12. A polynomial $f(x) = x^3 + ax^2 + bx + c$ has factors $(x-1)$, $(x+2)$, and $(x-3)$. What is the value of the constant term $c$?
    Answer: $c = 6$ When you see a polynomial with known factors, you're working with the relationship between roots and coefficients. Since the polynomial has factors $(x-1)$, $(x+2)$, and $(x-3)$, you can write it as $f(x) = (x-1)(x+2)(x-3)$. To find the constant term $c$, expand this factored form. Start by multiplying the first two factors: $(x-1)(x+2) = x^2 + 2x - x - 2 = x^2 + x - 2$. Next, multiply this result by the third factor: $(x^2 + x - 2)(x-3)$. When you expand this completely, you get: $x^3 - 3x^2 + x^2 - 3x - 2x + 6 = x^3 - 2x^2 - 5x + 6$ Therefore, $c = 6$, making choice C correct. There's also a shortcut: for any polynomial with roots $r_1, r_2, r_3$, the constant term equals $(-1)^n \cdot r_1 \cdot r_2 \cdot r_3$ where $n$ is the degree. Here, the roots are 1, -2, and 3, so $c = (-1)^3 \cdot 1 \cdot (-2) \cdot 3 = -1 \cdot (-6) = 6$. Choice A ($c = 0$) would mean one root is zero, which contradicts the given factors. Choice B ($c = -6$) results from forgetting the $(-1)^n$ factor in the shortcut formula. Choice D ($c = 1$) might come from incorrectly multiplying just two of the three roots. Remember: when you know all factors of a polynomial, you can always find any coefficient by expanding completely or using the relationship between roots and coefficients.
13. The polynomial $4x^2 - 12x + 9$ can be written in the form $(ax + b)^2$. What are the values of $a$ and $b$?
    Answer: $a = 2, b = -3$ When you see a polynomial that needs to be written as a perfect square, you're working with factoring—specifically recognizing or creating perfect square trinomials of the form $(ax + b)^2 = a^2x^2 + 2abx + b^2$. To factor $4x^2 - 12x + 9$, you can either expand each answer choice or work backwards from the polynomial's structure. Let's use the expansion method to check our answer systematically. The correct answer is C: $a = 2, b = -3$. When we expand $(2x + (-3))^2 = (2x - 3)^2$, we get: $(2x - 3)^2 = (2x)^2 + 2(2x)(-3) + (-3)^2 = 4x^2 - 12x + 9$ This matches our original polynomial exactly. Let's check why the other options fail: - A) $(4x - 3)^2 = 16x^2 - 24x + 9$ — the $x^2$ coefficient is wrong (16 instead of 4) - B) $(2x + 3)^2 = 4x^2 + 12x + 9$ — the middle term has the wrong sign (+12x instead of -12x) - D) $(-2x + 3)^2 = 4x^2 - 12x + 9$ — this actually gives the same result as C, but the conventional form uses positive leading coefficients Strategy tip: When factoring perfect square trinomials, always check that your factored form expands back to the original. Also remember that $(ax + b)^2$ and $(-ax - b)^2$ give the same result, but choose the form with a positive leading coefficient unless specified otherwise.
14. The polynomial $6x^3 + 11x^2 - 10x$ can be written as $x(2x + 5)(3x + k)$ for some integer $k$. What is the value of $k$?
    Answer: $-2$ This question tests polynomial factoring, specifically your ability to work backwards from a given factored form to find unknown coefficients. When you have a polynomial that's already partially factored like $x(2x + 5)(3x + k)$, you can find the missing value by expanding this expression and matching it to the original polynomial $6x^3 + 11x^2 - 10x$. Let's expand $x(2x + 5)(3x + k)$. First, multiply the binomials: $(2x + 5)(3x + k) = 6x^2 + 2kx + 15x + 5k = 6x^2 + (2k + 15)x + 5k$. Now multiply by $x$: $x[6x^2 + (2k + 15)x + 5k] = 6x^3 + (2k + 15)x^2 + 5kx$. Comparing coefficients with $6x^3 + 11x^2 - 10x$: - $x^3$ coefficient: $6 = 6$ ✓ - $x^2$ coefficient: $2k + 15 = 11$, so $2k = -4$, giving $k = -2$ - $x$ coefficient: $5k = 5(-2) = -10$ ✓ Choice A ($k = 2$) would give $2k + 15 = 19$, not 11. Choice C ($k = -5$) would give $2k + 15 = 5$, not 11. Choice D ($k = 5$) would give $2k + 15 = 25$, not 11. Therefore, $k = -2$, which is choice B. Study tip: When finding unknown coefficients in factored forms, expand the given expression and match coefficients term by term. This systematic approach prevents arithmetic errors and works for any polynomial degree.
15. The expression $x^3 - 8$ can be factored using the difference of cubes pattern. What is the complete factorization?
    Answer: $(x - 2)(x^2 + 2x + 4)$ Using $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ with $a = x$ and $b = 2$: $x^3 - 8 = (x-2)(x^2 + 2x + 4)$. Choice B uses the wrong middle term in the trinomial. Choice C has $4x$ instead of $2x$. Choice D uses the wrong sign in the first factor.
16. A student attempts to factor $2x^3 - 8x^2 + 6x$ and writes $2x(x^2 - 4x + 3)$. To complete the factorization, what should the student write next?
    Answer: $2x(x - 1)(x - 3)$ When you encounter a polynomial factoring problem, you're looking to break down the expression into its simplest multiplicative components. The student has correctly factored out the greatest common factor of $2x$, leaving $2x(x^2 - 4x + 3)$. Now you need to factor the quadratic expression $x^2 - 4x + 3$. To factor $x^2 - 4x + 3$, you need two numbers that multiply to give the constant term (3) and add to give the coefficient of the middle term (-4). Those numbers are -1 and -3, since $(-1) \times (-3) = 3$ and $(-1) + (-3) = -4$. Therefore, $x^2 - 4x + 3 = (x - 1)(x - 3)$, making the complete factorization $2x(x - 1)(x - 3)$. Choice A is incorrect because while it attempts the right format, the factors $(x - 4)(x + 1)$ would expand to $x^2 - 3x - 4$, not $x^2 - 4x + 3$. Choice B is wrong because it's not even in proper factored form—it contains subtraction outside the parentheses and would give an entirely different expression when expanded. Choice D is incorrect because the quadratic $x^2 - 4x + 3$ can indeed be factored further, so the factorization is not complete. Remember: always check if quadratic expressions can be factored further by looking for two numbers that multiply to the constant term and add to the linear coefficient. Complete factorization means breaking down until no factor can be simplified more.
17. The expression $x^4 - 16$ can be factored completely over the real numbers. How many linear factors does the complete factorization contain?
    Answer: 2 linear factors When you encounter an expression like $x^4 - 16$, you're looking at a difference of squares that can be factored step by step. The key is recognizing that this can be written as $(x^2)^2 - 4^2$, which follows the pattern $a^2 - b^2 = (a+b)(a-b)$. First, factor $x^4 - 16$ as a difference of squares: $(x^2)^2 - 4^2 = (x^2 + 4)(x^2 - 4)$. But you're not done yet! The factor $(x^2 - 4)$ is also a difference of squares that can be factored further: $x^2 - 4 = (x+2)(x-2)$. However, $(x^2 + 4)$ cannot be factored over the real numbers since it would require imaginary solutions. The complete factorization is: $x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)$. This gives us exactly 2 linear factors: $(x + 2)$ and $(x - 2)$. Choice A (0 linear factors) ignores that difference of squares creates linear terms. Choice B (3 linear factors) might come from incorrectly trying to factor $x^2 + 4$ over the reals. Choice C (4 linear factors) assumes you can factor $x^2 + 4$ into two linear factors, which requires complex numbers, not real numbers. Remember: when factoring completely over the real numbers, stop when you reach quadratic factors with no real roots (like $x^2 + 4$). Always check if your factors can be factored further before counting linear terms.
18. If $x^2 - kx + 16 = (x - m)^2$ for some constants $k$ and $m$, what is the value of $k$?
    Answer: $8$ If $x^2 - kx + 16 = (x - m)^2$, then expanding the right side gives $x^2 - 2mx + m^2$. Comparing coefficients: the constant term gives $m^2 = 16$, so $m = ±4$. The coefficient of $x$ gives $-k = -2m$, so $k = 2m = 2|m| = 8$. We can verify: $(x - 4)^2 = x^2 - 8x + 16$ or $(x + 4)^2 = x^2 + 8x + 16$. Since the original has $-kx$, we need $k = 8$.
19. Which expression is the complete factorization of $16x^4 - 81y^4$?
    Answer: $(2x - 3y)(2x + 3y)(4x^2 + 9y^2)$ $16x^4 - 81y^4 = (4x^2)^2 - (9y^2)^2 = (4x^2 - 9y^2)(4x^2 + 9y^2)$. This is a difference of squares pattern. Choice B has incorrect first factors. Choice C has wrong coefficients. Choice D treats it as if the variables had power 1.
20. If $p(x) = 6x^3 - 24x^2 + 18x$ and $p(x) = 6x \cdot q(x)$, what is the factored form of $q(x)$?
    Answer: $(x - 1)(x - 3)$ First, factor out the GCF from $p(x) = 6x^3 - 24x^2 + 18x = 6x(x^2 - 4x + 3)$. Since $p(x) = 6x \cdot q(x)$, we have $q(x) = x^2 - 4x + 3$. To factor this trinomial, we need two numbers that multiply to 3 and add to -4. These are -1 and -3, so $q(x) = (x - 1)(x - 3)$. Choice B uses incorrect factors, choice C leaves the quadratic unfactored, and choice D has wrong signs.

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