Polynomial Zeros - Accuplacer Aaf

What is the product of the zeros of the polynomial function $f(x) = 2x^3 - 5x^2 + 8x - 7$ ?
Answer: $\frac{7}{2}$ For a polynomial of the form $ax^3 + bx^2 + cx + d$ , the product of the zeros is given by the formula $-\frac{d}{a}$ . In this polynomial, $a = 2$ and $d = -7$ . Therefore, the product of the zeros is $-\frac{-7}{2}$ = $\frac{7}{2}$ .
The polynomial $f(x) = x^3 - 6x^2 + 11x - 6$ can be factored as $(x-1)(x-2)(x-3)$ . If $g(x) = f(x-2)$ , what are the zeros of $g(x)$ ?
Answer: $x = 3, x = 4, x = 5$ (shifted right by 2) If $f(x) = (x-1)(x-2)(x-3)$ has zeros at $x = 1, 2, 3$ , then $g(x) = f(x-2)$ represents a horizontal shift right by 2 units. To find zeros of $g(x)$ , solve $g(x) = f(x-2) = 0$ , which means $f(x-2) = 0$ . This occurs when $x-2 = 1, 2, 3$ , so $x = 3, 4, 5$ . Choice A incorrectly shifts left. Choice B ignores the transformation. Choice D incorrectly applies reflection.
If $r$ and $s$ are the zeros of a polynomial $P(x)$ , which of the following polynomials must have zeros at $r-2$ and $s-2$ ?
Answer: $P(x+2)$ Let $g(x)$ be the new polynomial. We want the zeros of $g(x)$ to be $r-2$ and $s-2$ . Let $z$ be a zero of $g(x)$ . Then $z = r-2$ or $z = s-2$ . This implies $z+2 = r$ or $z+2 = s$ . Since $r$ and $s$ are zeros of $P(x)$ , it means that $P(r)=0$ and $P(s)=0$ . Therefore, $P(z+2)=0$ . The new polynomial is $g(x) = P(x+2)$ .
A polynomial with rational coefficients has $2 - \sqrt{3}$ as one of its zeros. Which of the following must also be a zero of this polynomial?
Answer: $2 + \sqrt{3}$ The Irrational Conjugate Root Theorem states that if a polynomial has rational coefficients and an irrational zero of the form $a+b\sqrt{c}$ , then its conjugate $a-b\sqrt{c}$ must also be a zero. The conjugate of $2 - \sqrt{3}$ is $2 + \sqrt{3}$ .
If $r(x)$ is a polynomial function and the equation $r(x) = 5$ has solutions $x = 2$ and $x = 6$ , which statement must be true about $r(x) - 5$ ?
Answer: $r(x) - 5$ has zeros at $x = 2$ and $x = 6$ , and possibly others If $r(x) = 5$ when $x = 2$ and $x = 6$ , then $r(2) - 5 = 0$ and $r(6) - 5 = 0$ . This means $r(x) - 5$ has zeros at $x = 2$ and $x = 6$ . However, since we don't know the degree of $r(x)$ , there could be additional zeros of $r(x) - 5$ . Choice A incorrectly assumes these are the only zeros. Choice C confuses the input and output values. Choice D misunderstands that $r(x) = 5$ having solutions means $r(x) - 5$ has zeros.
What is the product of the real zeros of the equation $x^4 - 13x^2 + 36 = 0$ ?
Answer: 36 This equation is quadratic in form. Let $u = x^2$ . The equation becomes $u^2 - 13u + 36 = 0$ , which factors as $(u-9)(u-4) = 0$ . So, $u=9$ or $u=4$ . Substituting back, we have $x^2 = 9$ or $x^2 = 4$ . The solutions are $x = \pm 3$ and $x = \pm 2$ . All four zeros are real. Their product is $(3)(-3)(2)(-2) = (-9)(4) = 36$ .
For what value of $k$ is $(x-2)$ a factor of the polynomial $P(x) = x^3 - 3x^2 + kx - 10$ ?
Answer: 7 According to the Factor Theorem, if $(x-2)$ is a factor of $P(x)$ , then $P(2)$ must equal 0. Substitute $x=2$ into the polynomial: $P(2) = (2)^3 - 3(2)^2 + k(2) - 10 = 8 - 3(4) + 2k - 10 = 8 - 12 + 2k - 10$ . Simplifying gives $-14 + 2k = 0$ . Solving for $k$ yields $2k = 14$ , so $k = 7$ .
How many distinct real zeros does the polynomial function $f(x) = (x^2 + 9)(x-5)^2(x+1)$ have?
Answer: 2 To find the real zeros, set each factor to zero. The factor $(x^2 + 9)$ gives $x^2 = -9$ , which has no real solutions (only complex solutions $\pm 3i$ ). The factor $(x-5)^2$ gives $x-5=0$ , so $x=5$ is one distinct real zero. The factor $(x+1)$ gives $x+1=0$ , so $x=-1$ is a second distinct real zero. In total, there are 2 distinct real zeros.
Given that $x=1$ is a zero of $f(x) = x^3 - x^2 - 9x + 9$ , what is the sum of the other two zeros?
Answer: 0 For a cubic polynomial $ax^3+bx^2+cx+d$ , the sum of the three zeros is $-\frac{b}{a}$ . Here, $a=1$ and $b=-1$ , so the sum of all three zeros is $-\frac{-1}{1} = 1$ . If one zero is 1, and the other two are $r_2$ and $r_3$ , then $1 + r_2 + r_3 = 1$ . Subtracting 1 from both sides gives $r_2 + r_3 = 0$ .
The graph of the polynomial $P(x) = (x-a)^2 (x-1)$ passes through the point $(0, -16)$ . If $a > 0$ , what is the value of $a$ ?
Answer: 4 The point $(0, -16)$ is the y-intercept, which means that $P(0) = -16$ . Substitute $x=0$ into the polynomial's equation: $P(0) = (0-a)^2(0-1) = (-a)^2(-1) = -a^2$ . Setting this equal to -16 gives $-a^2 = -16$ , which simplifies to $a^2 = 16$ . The possible values for $a$ are 4 and -4. Since the problem states that $a > 0$ , the value of $a$ must be 4.
The graph of a polynomial function $y = f(x)$ is tangent to the x-axis at $x=-4$ and crosses the x-axis at $x=1$ . Which of the following could be the equation for $f(x)$ ?
Answer: $f(x) = (x+4)^2(x-1)$ If the graph is tangent to the x-axis at a zero, that zero has an even multiplicity (e.g., 2, 4, ...). Tangency at $x=-4$ implies a factor of $(x+4)^k$ where $k$ is even. If the graph crosses the x-axis, the zero has an odd multiplicity (e.g., 1, 3, ...). Crossing at $x=1$ implies a factor of $(x-1)^m$ where $m$ is odd. The simplest form matching this is $(x+4)^2(x-1)$ .
A polynomial function $P(x)$ has zeros at $x=-1$ and $x=5$ . Which of the following functions, $g(x)$ , must have zeros at $x=2$ and $x=8$ ?
Answer: $g(x) = P(x-3)$ Let the new zeros be $x_n$ . We want $g(x_n) = 0$ . For $g(x) = P(x-3)$ , we have $P(x_n-3) = 0$ . This means the input to $P$ , which is $x_n-3$ , must be one of the original zeros (-1 or 5). If $x_n-3 = -1$ , then $x_n = 2$ . If $x_n-3 = 5$ , then $x_n = 8$ . These are the desired zeros.
A polynomial is defined by $P(x) = x^2(x-3)(x+2)^3$ . What is the sum of the distinct real zeros of $P(x)$ ?
Answer: 1 The zeros of a polynomial are the values of $x$ for which $P(x) = 0$ . The factors are $x^2$ , $(x-3)$ , and $(x+2)^3$ . Setting each factor to zero gives $x=0$ , $x=3$ , and $x=-2$ . These are the distinct real zeros. The sum is $0 + 3 + (-2) = 1$ .
The polynomial $f(x) = x^3 - 2x^2 + kx + 18$ has a zero at $x=-3$ . What is the value of $k$ ?
Answer: -9 If $x=-3$ is a zero of the polynomial, then $f(-3) = 0$ . Substitute $x=-3$ into the function: $f(-3) = (-3)^3 - 2(-3)^2 + k(-3) + 18 = 0$ . This becomes $-27 - 2(9) - 3k + 18 = 0$ , which simplifies to $-27 - 18 - 3k + 18 = 0$ , or $-27 - 3k = 0$ . Solving for $k$ gives $-3k = 27$ , so $k = -9$ .
According to the Rational Root Theorem, which of the following is NOT a possible rational zero of the polynomial $P(x) = 3x^4 - 5x^3 + x^2 - 8x + 4$ ?
Answer: $\frac{3}{2}$ The Rational Root Theorem states that any possible rational zero must be of the form $\frac{p}{q}$ , where $p$ is a factor of the constant term (4) and $q$ is a factor of the leading coefficient (3). The factors of 4 are $\pm 1, \pm 2, \pm 4$ . The factors of 3 are $\pm 1, \pm 3$ . The value $\frac{3}{2}$ is not possible because 3 is not a factor of 4 and 2 is not a factor of 3.
A cubic polynomial has integer coefficients. If two of its zeros are $x=5$ and $x=\sqrt{2}$ , which of the following must also be a zero of the polynomial?
Answer: $-\sqrt{2}$ The Irrational Conjugate Root Theorem states that if a polynomial with rational (and thus integer) coefficients has an irrational zero of the form $a + b\sqrt{c}$ , then its conjugate $a - b\sqrt{c}$ must also be a zero. In this case, the zero $\sqrt{2}$ can be written as $0 + \sqrt{2}$ . Its conjugate is $0 - \sqrt{2}$ , or $-\sqrt{2}$ .
If $p(x) = 2x^4 - 8x^3 + 6x^2 + 4x - 4$ and $p(1) = 0$ , which statement about the polynomial $q(x) = \frac{p(x)}{x-1}$ is most accurate?
Answer: $q(x)$ is a cubic polynomial with the same zeros as $p(x)$ except $x = 1$ Since $p(1) = 0$ , we know $(x-1)$ is a factor of $p(x)$ , so $p(x) = (x-1) \cdot q(x)$ where $q(x)$ is a cubic polynomial. The zeros of $p(x)$ include $x = 1$ plus the zeros of $q(x)$ . Therefore $q(x)$ has the same zeros as $p(x)$ except for $x = 1$ . Choice B incorrectly suggests $q(x)$ still has $x = 1$ as a zero. Choice C misidentifies the degree. Choice D incorrectly suggests the division is undefined when it's actually valid polynomial division.
A polynomial with real coefficients has a zero of $4 - 5i$ . Which of the following must also be a zero of the polynomial?
Answer: $4 + 5i$ The Complex Conjugate Root Theorem states that if a polynomial has real coefficients and $a + bi$ is a zero, then its complex conjugate, $a - bi$ , must also be a zero. The complex conjugate of $4 - 5i$ is $4 + 5i$ .
A polynomial $P(x)$ has $P(-2) = 0$ and $P(5) = 0$ . Which of the following quadratic expressions must be a factor of $P(x)$ ?
Answer: $x^2 - 3x - 10$ If $P(-2)=0$ , then by the Factor Theorem, $(x - (-2))$ or $(x+2)$ is a factor. If $P(5)=0$ , then $(x-5)$ is a factor. Since both are factors, their product must also be a factor. The product is $(x+2)(x-5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10$ .
The zeros of the polynomial $P(x) = x^3 + kx^2 + 11x - 6$ are $r_1$ , $r_2$ , and $r_3$ . If $r_1 + r_2 + r_3 = 5$ , what is the value of $k$ ?
Answer: -5 For a cubic polynomial of the form $ax^3 + bx^2 + cx + d$ , the sum of the zeros is given by the formula $-\frac{b}{a}$ . In this polynomial, $a=1$ and $b=k$ . Therefore, the sum of the zeros is $-\frac{k}{1} = -k$ . We are given that the sum of the zeros is 5. So, $-k = 5$ , which means $k = -5$ .

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