1.
Question 1. Divide \((2x^4-5x^3+6x^2-5x+2)\) by \((x-2)\).
Step 1. Use synthetic division with \(x-2\), so the synthetic value is \(2\). Step 2. Use coefficients \(2,-5,6,-5,2\). Step 3. Synthetic division gives quotient coefficients \(2,-1,4,3\) and remainder \(8\). Step 4. Therefore, the quotient is \(2x^3-x^2+4x+3\) with remainder \(8\). Step 5. Write the result as \(2x^3-x^2+4x+3+\frac{8}{x-2}\). Answer: D. \(2x^3-x^2+4x+3+\frac{8}{x-2}\)
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Question 2. Find \(P(x)\) in the following equation: \(x^3+4x^2-1=P(x)(x+2)+7\).
Step 1. Subtract \(7\) from both sides: \(x^3+4x^2-8=P(x)(x+2)\). Step 2. Divide \(x^3+4x^2+0x-8\) by \(x+2\). Step 3. Use synthetic division with \(-2\) and coefficients \(1,4,0,-8\). Step 4. The quotient is \(x^2+2x-4\) and the remainder is \(0\). Step 5. Therefore, \(P(x)=x^2+2x-4\). Answer: C. \(x^2+2x-4\)
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Question 3. What is the remainder of \(6x^3+13x^2+x-2\) divided by \((x+2)\)?
Step 1. Use the Remainder Theorem. Step 2. For division by \(x+2\), substitute \(x=-2\). Step 3. Calculate \(P(-2)=6(-2)^3+13(-2)^2+(-2)-2\). Step 4. Simplify: \(P(-2)=-48+52-2-2=0\). Step 5. Therefore, the remainder is \(0\). Answer: A. \(0\)
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Question 4. What is the value of \(k\) if \(2x^3-4x^2+k-3\) is divided by \(x-1\) and gives a remainder of \(3\)?
Step 1. Use the Remainder Theorem. Step 2. For division by \(x-1\), substitute \(x=1\). Step 3. The remainder is \(P(1)=3\). Step 4. Calculate \(P(1)=2(1)^3-4(1)^2+k-3=2-4+k-3=k-5\). Step 5. Set \(k-5=3\). Step 6. Solve: \(k=8\). Answer: D. \(8\)
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Question 5. Using synthetic division, find the quotient \(Q(x)\) and the remainder \(R\) if \(P(x)=x^3-2x^2-5x+6\) is divided by \((x-1)\).
Step 1. Divide by \(x-1\), so use synthetic value \(1\). Step 2. Use coefficients \(1,-2,-5,6\). Step 3. Synthetic division gives quotient coefficients \(1,-1,-6\) and remainder \(0\). Step 4. Thus \(Q(x)=x^2-x-6\) and \(R=0\). Step 5. Since the remainder is \(0\), \((x-1)\) is a factor of \(P(x)\). Answer: D. \(Q(x)=x^2-x-6\) and \(R=0\); therefore, \((x-1)\) is a factor of \(P(x)\).
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Question 6. What is the factored form of \(x^3+x^2-5x+3\) if \((x-1)\) is a factor?
Step 1. Since \((x-1)\) is a factor, divide \(x^3+x^2-5x+3\) by \((x-1)\). Step 2. Synthetic division with \(1\) gives quotient \(x^2+2x-3\). Step 3. Factor the quotient: \(x^2+2x-3=(x+3)(x-1)\). Step 4. Combine factors: \((x-1)(x+3)(x-1)=(x-1)^2(x+3)\). Answer: B. \((x-1)^2(x+3)\)
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Question 7. Solve the polynomial equation \(x^3+3x^2-9x+5=0\).
Step 1. Test possible rational roots. Substitute \(x=1\): \(1+3-9+5=0\), so \(x=1\) is a root. Step 2. Divide \(x^3+3x^2-9x+5\) by \((x-1)\). Step 3. The quotient is \(x^2+4x-5\). Step 4. Factor the quotient: \(x^2+4x-5=(x+5)(x-1)\). Step 5. Therefore, the polynomial factors as \((x-1)^2(x+5)\). Step 6. The roots are \(x=1\) with multiplicity \(2\), and \(x=-5\). Answer: A. \(x=1\) with multiplicity \(2\), \(x=-5\).
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Question 8. Which of the following is not a root of the equation \(3x^3-2x^2-17x-12=0\)?
Step 1. Test option A by substituting \(x=2\). Step 2. Calculate \(3(2)^3-2(2)^2-17(2)-12=24-8-34-12=-30\). Step 3. Since the result is not \(0\), \(x=2\) is not a root. Step 4. The question asks which value is not a root. Answer: A. \(2\)
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Question 9. If \(P(x)=6x^3+13x^2+x-2\), and \(P(-2)=0\), then the factorization of \(P(x)\) is:
Step 1. Since \(P(-2)=0\), \((x+2)\) is a factor. Step 2. Divide \(6x^3+13x^2+x-2\) by \((x+2)\). Step 3. The quotient is \(6x^2+x-1\). Step 4. Factor the quotient: \(6x^2+x-1=(3x-1)(2x+1)\). Step 5. Therefore, \(P(x)=(x+2)(3x-1)(2x+1)\). Answer: A. \((x+2)(3x-1)(2x+1)\)
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Question 10. What are the zeros of the polynomial \((3x-1)(x+2)(2x+1)\)?
Step 1. Set each factor equal to \(0\). Step 2. \(3x-1=0\) gives \(x=\frac{1}{3}\). Step 3. \(x+2=0\) gives \(x=-2\). Step 4. \(2x+1=0\) gives \(x=-\frac{1}{2}\). Step 5. The zeros are \(-2,-\frac{1}{2},\frac{1}{3}\). Answer: B. \(-2,-\frac{1}{2},\frac{1}{3}\)
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Question 11. What is the equation of the polynomial function shown on the graph?
Step 1. From the graph, the zeros are approximately \(x=-2\), \(x=0\), and \(x=4\). Step 2. Therefore, the function must contain factors \((x+2)\), \(x\), and \((x-4)\). Step 3. The graph rises to the left and falls to the right, so the leading coefficient is negative. Step 4. The equation matching these features is \(y=-x(x+2)(x-4)\). Answer: B. \(y=-x(x+2)(x-4)\)
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Question 12. Which equation represents the polynomial function with zeros \(-5\), \(-\frac{1}{2}\), and \(3\), and a \(y\)-intercept of \(-15\)?
Step 1. A zero of \(-5\) gives the factor \((x+5)\). Step 2. A zero of \(-\frac{1}{2}\) gives the factor \((2x+1)\). Step 3. A zero of \(3\) gives the factor \((x-3)\). Step 4. Test the \(y\)-intercept by setting \(x=0\): \((0+5)(2(0)+1)(0-3)=5(1)(-3)=-15\). Step 5. This matches the required \(y\)-intercept. Answer: B. \(y=(x+5)(2x+1)(x-3)\)
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Question 13. Which of the following represents the graph of \(y=-(x+3)^2(2x-1)(x-4)\)?
Step 1. Identify the zeros: \(x=-3\), \(x=\frac{1}{2}\), and \(x=4\). Step 2. The zero \(x=-3\) has multiplicity \(2\), so the graph touches the \(x\)-axis at \(-3\) and turns around. Step 3. The other zeros have odd multiplicity, so the graph crosses the \(x\)-axis at \(\frac{1}{2}\) and \(4\). Step 4. The degree is \(4\) and the leading coefficient is negative, so both ends of the graph point downward. Step 5. The graph matching these features is Graph D. Answer: D. Graph D
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Question 14. Which of the following is a polynomial with a degree of \(5\), has \(3\) distinct zeros, has \(2\) zeros with a multiplicity of \(2\), and has a positive leading coefficient?
Step 1. A degree \(5\) polynomial has odd-degree end behavior. Step 2. A positive leading coefficient means the graph goes down on the left and up on the right. Step 3. Having \(3\) distinct zeros with \(2\) zeros of multiplicity \(2\) means two zeros should touch and turn, while one zero should cross. Step 4. Graph D matches this behavior and these zero multiplicities. Answer: D. Graph D
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Question 15. What is the value of the leading coefficient \(a\) if the polynomial function \(P(x)=a(x+b)^2(x-c)\) has multiplicity of \(2\) at the point \((-3,0)\) and also passes through the points \((2,0)\) and \((0,36)\)?
Step 1. A zero with multiplicity \(2\) at \((-3,0)\) gives factor \((x+3)^2\). Step 2. A zero at \((2,0)\) gives factor \((x-2)\). Step 3. The function is \(P(x)=a(x+3)^2(x-2)\). Step 4. Use point \((0,36)\): \(36=a(0+3)^2(0-2)\). Step 5. Simplify: \(36=a(9)(-2)=-18a\). Step 6. Solve: \(a=-2\). Answer: A. \(-2\)
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Question 16. Which rational function has zeros at \(x=1\) and \(x=3\)?
Step 1. Zeros of a rational function come from the numerator, as long as they are not cancelled by the denominator. Step 2. Factor option B numerator: \(2x^2-8x+6=2(x^2-4x+3)=2(x-1)(x-3)\). Step 3. Thus the numerator is zero at \(x=1\) and \(x=3\). Step 4. The denominator \(x^2-4x+4=(x-2)^2\), so it does not cancel \((x-1)\) or \((x-3)\). Answer: B. \(f(x)=\frac{2x^2-8x+6}{x^2-4x+4}\)
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Question 17. For the function defined by \(f(x)=\frac{4x^2+12x}{x^2-9}\), what is occurring at \(x=-3\)?
Step 1. Factor the numerator: \(4x^2+12x=4x(x+3)\). Step 2. Factor the denominator: \(x^2-9=(x-3)(x+3)\). Step 3. The factor \((x+3)\) cancels, so there is a hole at \(x=-3\), not a vertical asymptote. Step 4. The simplified function is \(f(x)=\frac{4x}{x-3}\). Step 5. Find the hole's \(y\)-value by substituting \(x=-3\): \(f(-3)=\frac{4(-3)}{-3-3}=\frac{-12}{-6}=2\). Step 6. Therefore, there is a hole at \((-3,2)\). Answer: C. The graph of \(f\) has a hole at \((-3,2)\).
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Question 18. Find the domain of the rational function \(f(x)=\frac{2x^2-8x+6}{x^2-4x+4}\).
Step 1. The domain of a rational function excludes values that make the denominator equal to \(0\). Step 2. Factor the denominator: \(x^2-4x+4=(x-2)^2\). Step 3. Set the denominator equal to \(0\): \((x-2)^2=0\). Step 4. Solve: \(x=2\). Step 5. Therefore, the domain is all real numbers except \(x=2\). Answer: B. Domain: all real numbers except \(x=2\)
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