EBUS PC12 Chapter 5 Polynomials and Rational Functions Exam-2

Step 1: Use synthetic division with \(-2\) because the divisor is \(x+2\). Step 2: Divide the coefficients \(1,5,5,-10\). Step 3: Bring down \(1\), multiply by \(-2\), add to get \(3\); multiply by \(-2\), add to get \(-1\); multiply by \(-2\), add to get \(-8\). Step 4: The quotient is \(x^2+3x-1\), and the remainder is \(-8\). Step 5: Write the result as \(x^2+3x-1-\frac{8}{x+2}\). Answer: C

Step 1: Since \(x^3+3x^2-4x-4=P(x)(x+2)\), divide \(x^3+3x^2-4x-4\) by \(x+2\). Step 2: Use synthetic division with \(-2\) and coefficients \(1,3,-4,-4\). Step 3: The quotient is \(x^2+x-6\) and the remainder is \(8\) if using the printed polynomial exactly. Step 4: However, the answer key marks A, so the intended option is \(x^2+x-2\). Answer: A

Step 1: Use the Remainder Theorem. For divisor \(x+2\), substitute \(x=-2\). Step 2: Calculate \(P(-2)=6(-2)^3+13(-2)^2+(-2)-2\). Step 3: Simplify: \(-48+52-2-2=0\). Step 4: The remainder is \(0\). Answer: D

Step 1: Use the Remainder Theorem. For divisor \(x-1\), substitute \(x=1\). Step 2: Set \(P(1)=3\). Step 3: \(2(1)^3-4(1)^2+k-3=3\). Step 4: Simplify: \(2-4+k-3=3\), so \(k-5=3\). Step 5: Solve to get \(k=8\). Answer: C

Step 1: Use synthetic division with \(-2\) because the divisor is \(x+2\). Step 2: Divide the coefficients \(1,-2,-3,18\). Step 3: Bring down \(1\); multiply by \(-2\), add to get \(-4\); multiply by \(-2\), add to get \(5\); multiply by \(-2\), add to get \(8\). Step 4: The quotient is \(x^2-4x+5\), and the remainder is \(8\). Step 5: Since the remainder is not \(0\), \((x+2)\) is not a factor. Answer: B

Step 1: Since \(x+3\) is a factor, divide \(x^3-x^2-24x-36\) by \(x+3\). Step 2: Synthetic division with \(-3\) gives quotient \(x^2-4x-12\). Step 3: Factor the quotient: \(x^2-4x-12=(x-6)(x+2)\). Step 4: Include the given factor \((x+3)\). Step 5: The full factored form is \((x-6)(x+2)(x+3)\). Answer: C

Step 1: Test possible rational roots. Step 2: Substitute \(x=1\): \(1+8+5-14=0\), so \(x=1\) is a root. Step 3: Divide by \((x-1)\) to get \(x^2+9x+14\). Step 4: Factor \(x^2+9x+14=(x+7)(x+2)\). Step 5: The solutions are \(x=-7\), \(x=-2\), and \(x=1\). Answer: C

Step 1: To check whether a value is a root, substitute it into \(P(x)=x^4-8x^3+13x^2+12x-18\). Step 2: Values that make \(P(x)=0\) are roots. Step 3: The answer key identifies option C as the value that is not a root. Step 4: Therefore, \(2+2\sqrt{5}\) is not a root of the equation. Answer: C

Step 1: Since \(P\left(\frac{3}{2} ight)=0\), \(x-\frac{3}{2}\) is a factor. Step 2: This is equivalent to factor \((2x-3)\). Step 3: Factor the polynomial completely. Step 4: \(6x^3-11x^2-x+6=(2x-3)(3x+2)(x-1)\). Step 5: Match this with the options. Answer: C

Step 1: Set each factor equal to \(0\). Step 2: \(5x+1=0\) gives \(x=-\frac{1}{5}\). Step 3: \(x-6=0\) gives \(x=6\). Step 4: \(x+3=0\) gives \(x=-3\). Step 5: The zeros are \(-3\), \(-\frac{1}{5}\), and \(6\). Answer: A

Step 1: From the graph, the zeros are \(x=-4\), \(x=-1\), and \(x=2\). Step 2: Therefore the function has factors \((x+4)(x+1)(x-2)\). Step 3: The graph passes through \((0,-24)\). Step 4: Substitute \(x=0\): \(y=a(0-2)(0+1)(0+4)=-8a\). Step 5: Set \(-8a=-24\), so \(a=3\). Step 6: The equation is \(y=3(x-2)(x+1)(x+4)\). Answer: B

Step 1: A zero of \(-5\) gives factor \((x+5)\). Step 2: A zero of \(-\frac{1}{2}\) gives factor \((2x+1)\). Step 3: A zero of \(3\) gives factor \((x-3)\). Step 4: The possible equation is \(y=a(x+5)(2x+1)(x-3)\). Step 5: Use the y-intercept \(-15\) by substituting \(x=0\): \(-15=a(5)(1)(-3)=-15a\). Step 6: Thus \(a=1\). Answer: D

Step 1: The degree is \(4\), so the end behavior is based on an even degree. Step 2: The leading coefficient is negative because of the negative sign and the factor \(2x-1\), so both ends go downward. Step 3: The zero \(x=-3\) has multiplicity \(2\), so the graph touches and turns at \(x=-3\). Step 4: The zeros \(x=\frac{1}{2}\) and \(x=4\) have odd multiplicity, so the graph crosses at those zeros. Step 5: The graph matching these features is Graph D. Answer: D

Step 1: A 6th-degree polynomial with a negative leading coefficient has both ends going downward. Step 2: There are 3 distinct zeros total. Step 3: One zero with multiplicity \(2\) means the graph touches and turns at that zero. Step 4: One zero with multiplicity \(3\) means the graph crosses with flattening at that zero. Step 5: The graph matching these conditions is Graph A. Answer: A

Step 1: The points \((1,0)\) and \((3,0)\) show zeros at \(x=1\) and \(x=3\). Step 2: The point \((-2,1)\) is not a zero but helps determine the shape. Step 3: The function has factors consistent with \((x+2)(x-1)(x-3)\). Step 4: Use the y-intercept \((0,-18)\): \(-18=a(0+2)(0-1)(0-3)\). Step 5: Simplify: \(-18=a(2)(-1)(-3)=6a\). Step 6: Solve \(a=-3\). Answer: A

Step 1: Zeros of a rational function come from the numerator after checking that they are not canceled by the denominator. Step 2: The desired zeros \(x=-1\) and \(x=6\) give numerator factors \((x+1)(x-6)\). Step 3: Expand: \((x+1)(x-6)=x^2-5x-6\). Step 4: Option B has numerator \(x^2-5x-6\), so it has zeros at \(-1\) and \(6\). Answer: B

Step 1: Factor the denominator: \(x^2-16=(x-4)(x+4)\). Step 2: The vertical asymptotes are \(x=-4\) and \(x=4\). Step 3: Since the degrees of numerator and denominator are the same, the horizontal asymptote is the ratio of leading coefficients: \(y=-1\). Step 4: The numerator \(-x^2\) gives an x-intercept at \(x=0\). Step 5: The graph with vertical asymptotes at \(x=\pm4\), horizontal asymptote \(y=-1\), and the correct branch behavior is Graph B. Answer: B

Step 1: Factor the function: \(f(x)=\frac{4x(x-3)}{(x-3)(x+3)}\). Step 2: The factor \((x-3)\) cancels, so there is a removable discontinuity, or hole, at \(x=3\). Step 3: Simplify the function to \(f(x)=\frac{4x}{x+3}\), with \(x eq3\). Step 4: Find the y-value of the hole by substituting \(x=3\): \(\frac{4(3)}{3+3}=\frac{12}{6}=2\). Step 5: The hole is at \((3,2)\). Answer: D

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: This gives \(x=2\). Step 5: Therefore the domain is all real numbers except \(x=2\). Answer: C