Factor Theorem - Math 3

The polynomial $K(x) = x^5 - 3x^4 - 13x^3 + 39x^2 + 36x - 108$ has $(x - 3)$ as a factor. After factoring out $(x - 3)$ , the resulting quotient polynomial $Q(x)$ also has $(x - 3)$ as a factor. What is the minimum degree of multiplicity of the root $x = 3$ in $K(x)$ ?
Answer: The root $x = 3$ has multiplicity at least 2, but could be higher without further testing The given information tells us that $K(x) = (x-3) \cdot Q(x)$ and that $Q(x)$ also has $(x-3)$ as a factor, meaning $Q(x) = (x-3) \cdot R(x)$ for some polynomial $R(x)$ . Therefore $K(x) = (x-3)^2 \cdot R(x)$ , so the multiplicity is at least 2. However, without additional information (such as testing whether $R(3) = 0$ ), we cannot determine if the multiplicity is exactly 2 or higher. Choice A assumes exactly 2. Choice C assumes exactly 3 without justification. Choice D is incorrect since we can determine it's at least 2.
The polynomial $T(x) = 2x^4 + bx^3 + cx^2 + dx + e$ has the property that $T(1) = T(-1) = T(2) = T(-2) = 0$ . Which statement about the coefficients must be true?
Answer: The coefficients satisfy $b = d = 0$ and $c = -10, e = 8$ Since $T(x)$ has zeros at $±1$ and $±2$ , we have $T(x) = 2(x-1)(x+1)(x-2)(x+2) = 2(x^2-1)(x^2-4) = 2(x^4-5x^2+4) = 2x^4-10x^2+8$ . Comparing with $T(x) = 2x^4 + bx^3 + cx^2 + dx + e$ , we get $b = 0$ , $c = -10$ , $d = 0$ , and $e = 8$ . Choice A is partially correct but incomplete. Choice B is correct that they're uniquely determined but doesn't give the values. Choice C gives only $e$ correctly.
The polynomial $P(x) = 2x^3 + px^2 + qx - 12$ has $(x - 2)$ as a factor. Additionally, when $P(x)$ is divided by $(x + 1)$ , the remainder is $-15$ . What is the value of $p + q$ ?
Answer: $p + q = -1$ using the given constraints and polynomial division Since $(x - 2)$ is a factor, $P(2) = 0$ : $16 + 4p + 2q - 12 = 0$ , so $4p + 2q = -4$ or $2p + q = -2$ . Since the remainder when dividing by $(x + 1)$ is $-15$ , $P(-1) = -15$ : $-2 + p - q - 12 = -15$ , so $p - q = -1$ . Solving the system: $2p + q = -2$ and $p - q = -1$ gives $p = -1, q = 0$ . Therefore $p + q = -1$ .
Given that $T(x) = 2x^3 - 7x^2 + 4x + k$ where $k$ is a constant, and $(x-2)$ is a factor of $T(x)$ , what is the remainder when $T(x+1)$ is divided by $(x-1)$ ?
Answer: The remainder is $0$ because the translation preserves the factor relationship First, since $(x-2)$ is a factor of $T(x)$ , we have $T(2) = 0$ . This gives us $T(2) = 2(8) - 7(4) + 4(2) + k = 16 - 28 + 8 + k = -4 + k = 0$ , so $k = 4$ . Therefore, $T(x) = 2x^3 - 7x^2 + 4x + 4$ . Now, $T(x+1) = 2(x+1)^3 - 7(x+1)^2 + 4(x+1) + 4$ . By the Remainder Theorem, the remainder when $T(x+1)$ is divided by $(x-1)$ is $T(1+1) = T(2) = 0$ . This makes sense because if $(x-2)$ is a factor of $T(x)$ , then $T(x+1)$ has $(x-1)$ as a factor (since when $x = 1$ , we get $x+1 = 2$ , which is a zero of $T$ ).
Given that $P(x) = 2x^4 - 5x^3 - 23x^2 + 60x$ and $P(4) = 0$ , which of the following statements must be true?
Answer: $(x - 4)$ is a factor of $P(x)$ , and $P(x) = (x - 4)(2x^3 - 13x^2 + 29x - 15)$ By the Factor Theorem, if $P(4) = 0$ , then $(x - 4)$ is a factor of $P(x)$ . Using polynomial long division or synthetic division: $P(x) = (x - 4)(2x^3 - 13x^2 + 29x - 15)$ . Choice A has the correct factor but wrong quotient. Choice B uses the wrong factor $(x + 4)$ . Choice D incorrectly assumes $(4x - 1)$ is a factor, which would require $P(1/4) = 0$ , not $P(4) = 0$ .
The polynomial $W(x) = x^3 + 2x^2 - 5x - 6$ can be factored as $(x + 1)(x + 3)(x - 2)$ . If $V(x) = W(2x)$ , which of the following represents the correct factorization of $V(x)$ ?
Answer: $V(x) = 8(x + \frac{1}{2})(x + \frac{3}{2})(x - 1)$ after scaling transformation When you encounter polynomial transformations like $V(x) = W(2x)$ , you're dealing with horizontal scaling that affects both the roots and the leading coefficient of the factored form. To find $V(x) = W(2x)$ , substitute $2x$ into the original polynomial: $V(x) = (2x)^3 + 2(2x)^2 - 5(2x) - 6 = 8x^3 + 8x^2 - 10x - 6$ . You can factor out 8: $V(x) = 8(x^3 + x^2 - \frac{5}{4}x - \frac{3}{4})$ . The key insight is understanding how the transformation affects the roots. If $W(x)$ has roots at $x = -1, -3, 2$ , then $V(x) = W(2x)$ has roots where $2x = -1, -3, 2$ , giving us $x = -\frac{1}{2}, -\frac{3}{2}, 1$ . The horizontal compression by factor 2 shrinks all root positions by half. Additionally, since the leading term becomes $8x^3$ , we get the factor of 8. Therefore, $V(x) = 8(x + \frac{1}{2})(x + \frac{3}{2})(x - 1)$ . Choice A ignores the transformation entirely. Choice B incorrectly substitutes $2x$ directly into each factor without considering how transformations actually affect roots. Choice C makes an error in the root calculations—these would correspond to $V(x) = W(\frac{x}{2})$ , not $V(x) = W(2x)$ . Remember: for $f(ax)$ transformations, divide the original roots by $a$ to find the new roots, and account for how the leading coefficient changes through the substitution.
The polynomial $f(x) = x^3 + ax^2 + bx + c$ has zeros at $x = -2$ , $x = 1$ , and $x = 5$ . If $g(x) = x^4 + ax^3 + bx^2 + cx + d$ and $g(-2) = 0$ , $g(1) = 0$ , $g(5) = 0$ , what can be concluded about the factorization of $g(x)$ ?
Answer: $g(x) = (x + 2)(x - 1)(x - 5)(x - k)$ for some value $k$ Since $g(-2) = g(1) = g(5) = 0$ , by the Factor Theorem, $(x + 2)$ , $(x - 1)$ , and $(x - 5)$ are all factors of $g(x)$ . Since $g(x)$ is degree 4, there must be exactly one more linear factor $(x - k)$ for some value $k$ . Choice B incorrectly restricts $k$ to be positive. Choice C assumes repeated roots without justification. Choice D is incorrect because $g(x) ≠ f(x) \cdot (x - k)$ since the coefficients don't match this pattern.
If $p(x) = x^4 - 5x^3 + 7x^2 - 5x + 6$ and $p(2) = 0$ , which statement about the polynomial $q(x) = x^3 - 3x^2 + x - 3$ is necessarily true?
Answer: $q(x)$ represents the quotient when $p(x)$ is divided by $(x-2)$ Since $p(2) = 0$ , by the Factor Theorem, $(x-2)$ is a factor of $p(x)$ . This means $p(x) = (x-2) \cdot q(x)$ where $q(x)$ is the quotient polynomial. Performing polynomial division of $p(x)$ by $(x-2)$ yields $q(x) = x^3 - 3x^2 + x - 3$ . Choice A is incorrect because $q(x)$ may factor further. Choice C is incorrect because $q(x)$ could have additional zeros or share some zeros with the original polynomial. Choice D is incorrect because there's no reason $q(x)$ would contain $(x-2)$ as a factor.
Consider two polynomials: $F(x) = x^3 - 6x^2 + 11x - 6$ and $G(x) = x^4 - 6x^3 + 11x^2 - 6x$ . If $F(1) = 0$ , $F(2) = 0$ , and $F(3) = 0$ , what can be concluded about the relationship between $F(x)$ and $G(x)$ using the Factor Theorem?
Answer: $G(x) = x \cdot F(x)$ but $G(x)$ has an additional root at $x = 0$ Since $F(1) = F(2) = F(3) = 0$ and $F(x)$ is cubic, we have $F(x) = (x-1)(x-2)(x-3)$ . Looking at $G(x) = x^4 - 6x^3 + 11x^2 - 6x = x(x^3 - 6x^2 + 11x - 6) = x \cdot F(x)$ . Therefore $G(x)$ has all the roots of $F(x)$ (namely 1, 2, 3) plus an additional root at $x = 0$ . Choice A is incomplete as it doesn't mention the additional root. Choice B gives the wrong relationship. Choice D is incorrect as the polynomials are clearly related.
Given that $P(x) = x^4 + mx^3 + nx^2 + px + q$ and that $P(x) - 12 = (x-2)^2(x+1)(x-3)$ , what is the value of $P(2)$ ?
Answer: $P(2) = 12$ since $P(x) - 12$ has $(x-2)^2$ as a factor Since $P(x) - 12 = (x-2)^2(x+1)(x-3)$ , we have $P(x) = (x-2)^2(x+1)(x-3) + 12$ . To find $P(2)$ , substitute $x = 2$ : $P(2) = (2-2)^2(2+1)(2-3) + 12 = 0 \cdot 3 \cdot (-1) + 12 = 0 + 12 = 12$ . Choice A incorrectly assumes $P(2) = 0$ . Choice C and D give incorrect values based on misunderstanding the relationship.
If $F(x) = x^4 - 10x^2 + 9$ and we know that $F(1) = 0$ , which statement about the complete factorization of $F(x)$ is correct?
Answer: $F(x) = (x-1)(x+1)(x-3)(x+3)$ revealing four distinct linear factors We can verify $F(1) = 1 - 10 + 9 = 0$ , confirming $(x-1)$ is a factor. Notice that $F(x) = x^4 - 10x^2 + 9$ is actually a quadratic in $x^2$ . Let $u = x^2$ , then $F(x) = u^2 - 10u + 9 = (u-1)(u-9) = (x^2-1)(x^2-9)$ . This can be factored further: $(x^2-1)(x^2-9) = (x-1)(x+1)(x-3)(x+3)$ . So the complete linear factorization has four factors, and we can verify that $x = \pm 1, \pm 3$ are all zeros. Choice A is incomplete, Choice C doesn't show complete linear factorization, and Choice D is incorrect as it suggests $(x-1)^2$ which would mean $F'(1) = 0$ , but $F'(x) = 4x^3 - 20x$ and $F'(1) = 4 - 20 = -16 \neq 0$ .
Let $Q(x) = x^4 - 10x^3 + 35x^2 - 50x + 24$ . If the Factor Theorem is used to test whether $(x - 2)$ is a factor, and it is found that $Q(2) = 0$ , what additional information is needed to determine if $(x - 2)^2$ is also a factor?
Answer: Calculate $Q'(2)$ and check if it equals zero for multiplicity determination The Factor Theorem only tells us that $(x - 2)$ is a factor when $Q(2) = 0$ . To determine if $(x - 2)^2$ is a factor (i.e., if $x = 2$ is a root of multiplicity at least 2), we need $Q'(2) = 0$ . If $Q'(2) = 0$ , then $x = 2$ has multiplicity ≥ 2. Choice B would work but is unnecessarily complicated. Choice C is irrelevant. Choice D tests for multiplicity ≥ 3, which goes beyond what's asked.
Consider the polynomial $h(x) = x^4 - 6x^3 + 9x^2 + 6x - 10$ . If it is known that $h(1) = 0$ and $h(5) = 0$ , which statement about the complete factorization is most likely correct?
Answer: $h(x) = (x - 1)(x - 5)(x^2 + 2)$ with no real linear factors remaining Using the Factor Theorem and polynomial division: $h(x) ÷ (x - 1)(x - 5) = h(x) ÷ (x^2 - 6x + 5) = x^2 + 2$ . Therefore $h(x) = (x - 1)(x - 5)(x^2 + 2)$ . The quadratic factor $x^2 + 2$ has no real roots. Choice B gives an incorrect factorization. Choice C has the wrong quadratic factor. Choice D assumes $x = 1$ is a repeated root, but $h'(1) ≠ 0$ .

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