Polynomial and Rational Functions Review: math test solutions, Grade 12 – download the PDF

1 End behavior and zeros / 3 pts

(a) Degree \(1+2+1=4\); leading coefficient \(-1\) (1 pt). Even degree and negative coefficient: the graph falls on both ends (0.5 pt).

(b) \(0\) (multiplicity 1, crosses), \(3\) (multiplicity 2, touches and turns), \(-1\) (multiplicity 1, crosses) (1.5 pts).

2 Synthetic division and factoring / 4 pts

(a) \(P(1)=1-2-5+6=0\) (1 pt).

(b)

Quotient \(x^2-x-6\), remainder \(0\) (1.5 pts).

(c) \(x^2-x-6=(x-3)(x+2)\), so \(P(x)=(x-1)(x-3)(x+2)\) with zeros \(1,3,-2\) (1.5 pts).

3 Remainder Theorem / 3 pts

(a) \(Q(2)=16-4+2k-4=8+2k=6\), so \(k=-1\) (1.5 pts).

(b) \(Q(x)=2x^3-x^2-x-4\) and \(Q(-1)=-2-1+1-4=-6\) (1.5 pts).

4 Real and complex zeros / 4 pts

(a) \(p\in\{\pm1,\pm2,\pm5,\pm10\}\), \(q\in\{1,2\}\): \(\pm1,\pm2,\pm5,\pm10,\pm\tfrac12,\pm\tfrac52\) (1 pt).

(b) \(R(-\tfrac12)=-\tfrac14-\tfrac{11}4-7+10=0\), so \(-\tfrac12\) is a zero (1 pt).

(c)

The quotient is \(2x^2-12x+20\), so \(R(x)=(x+\tfrac12)(2x^2-12x+20)=(2x+1)(x^2-6x+10)\) (1 pt).

(d) \(x^2-6x+10=0\) gives \(x=\dfrac{6\pm\sqrt{36-40}}{2}=3\pm i\). The zeros are \(-\tfrac12\), \(3+i\), and \(3-i\) (1 pt).

5 Asymptotes and intercepts / 3 pts

(a) \(x^2-x-6=(x-3)(x+2)\) and nothing cancels with \(4(x-2)\): vertical asymptotes \(x=3\) and \(x=-2\) (1 pt).

(b) The numerator has degree 1 and the denominator degree 2: \(y=0\) (1 pt).

(c) \(x\)-intercept: \(4x-8=0\), so \(x=2\). \(y\)-intercept: \(g(0)=\dfrac{-8}{-6}=\tfrac43\) (1 pt).

6 Rational inequality / 3 pts

Critical numbers: \(-2\) and \(3\) (excluded), \(2\) (included) (1 pt). Signs of \(g\) on \((-\infty,-2)\), \((-2,2)\), \((2,3)\), \((3,\infty)\): \(-,+,-,+\) (1 pt).

The solution is \((-2,2]\cup(3,\infty)\) (1 pt).