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

1 Degree and end behavior ★★★

The highest power is \(x^5\), so the degree is 5 and the leading coefficient is \(-2\). The \(y\)-intercept is \(f(0)=-7\).

The degree is odd and the leading coefficient is negative, so the graph rises on the left (\(f(x)\to+\infty\) as \(x\to-\infty\)) and falls on the right (\(f(x)\to-\infty\) as \(x\to+\infty\)).

2 Four end behaviors ★★★

1. Odd degree, positive leading coefficient: falls on the left, rises on the right.
2. Even degree, negative leading coefficient: falls on both ends.
3. Odd degree, negative leading coefficient: rises on the left, falls on the right.
4. Even degree, positive leading coefficient: rises on both ends.

3 Zeros and multiplicities ★★★

Set each factor to zero: \(x=0\) (multiplicity 2), \(x=4\) (multiplicity 1), \(x=-1\) (multiplicity 3).

At \(0\) the multiplicity is even, so the graph touches the axis and turns around. At \(4\) it crosses. At \(-1\) the multiplicity is odd, so it crosses while flattening out. The degree is \(2+1+3=6\).

4 A first synthetic division ★★★

Use \(c=2\) and coefficients \(1,-4,1,6\).

The quotient is \(x^2-2x-3\) and the remainder is \(0\). Therefore \(x^3-4x^2+x+6=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)\).

5 Remainders by evaluating ★★★

By the Remainder Theorem the remainder is \(P(c)\) for the divisor \(x-c\).

1. \(P(3)=27-15+4=16\).
2. The divisor is \(x-(-1)\), so the remainder is \(P(-1)=-1+5+4=8\).

6 Simple asymptotes ★★★

The denominator is zero at \(x=-4\) and the numerator is not, so the vertical asymptote is \(x=-4\). The degrees are equal (both 1), so the horizontal asymptote is the ratio of the leading coefficients: \(y=\tfrac31=3\).

7 A quick inequality ★★★

The critical numbers are \(-2\) and \(5\). The parabola opens upward, so the product is negative only between its zeros. Check \(x=0\): \((-5)(2)=-10\lt0\). The solution is \((-2,5)\).

8 True or false? ★★★

1. True. For example \(x^4+1\gt0\) for every real \(x\).
2. True. Non-real zeros come in conjugate pairs, so a cubic has 3 zeros with an even number of non-real ones (0 or 2). At least one zero is real.
3. False. Since \(f(x)=\dfrac{(x-1)(x+1)}{x-1}=x+1\) for \(x\ne1\), the factor cancels. There is a hole at \((1,2)\), not an asymptote.

9 Factor Theorem ★★★

\(P(-3)=-27+18+15-6=0\), so \(x+3\) is a factor by the Factor Theorem. Divide with \(c=-3\):

The quotient is \(x^2-x-2=(x-2)(x+1)\). So \(P(x)=(x+3)(x-2)(x+1)\).

10 Finding a missing coefficient ★★★

The Factor Theorem gives \(P(2)=8+4k-6-10=4k-8=0\), so \(k=2\) and \(P(x)=x^3+2x^2-3x-10\).

The quotient is \(x^2+4x+5\). Its discriminant is \(16-20=-4\lt0\), so \(x=\dfrac{-4\pm 2i}{2}=-2\pm i\). The zeros are \(2\), \(-2+i\), and \(-2-i\).

11 Rational Root Theorem at work ★★★

Divisors of the constant term \(6\): \(\pm1,\pm2,\pm3,\pm6\). Divisors of the leading coefficient \(2\): \(1,2\). Candidates: \(\pm1,\pm2,\pm3,\pm6,\pm\tfrac12,\pm\tfrac32\).

Test \(x=2\): \(16+4-26+6=0\). Divide:

The quotient is \(2x^2+5x-3=(2x-1)(x+3)\). The zeros are \(2\), \(\tfrac12\), and \(-3\).

12 An open-top box ★★★

1. The box has height \(x\) and a square base of side \(12-2x\), so \(V(x)=x(12-2x)^2\), with \(0\lt x\lt6\).
2. Solve \(x(12-2x)^2=128\). Expanding gives \(4x^3-48x^2+144x-128=0\), that is \(x^3-12x^2+36x-32=0\). Testing \(x=2\): \(8-48+72-32=0\). Synthetic division gives \(x^2-10x+16=(x-2)(x-8)\). The zeros are \(2\) and \(8\). Since \(x\) must lie in \((0,6)\), only \(x=2\) works. Check: base \(8\times8\), height \(2\), volume \(128\). So the corners are squares of side 2 inches.

13 Hole or asymptote? ★★★

\(f(x)=\dfrac{2(x-2)(x+2)}{(x-3)(x+2)}\). The factor \(x+2\) cancels, so there is a hole at \(x=-2\), at height \(\dfrac{2(-4)}{-5}=\tfrac85\).

The vertical asymptote is \(x=3\). The degrees are equal and \(\tfrac21=2\), so the horizontal asymptote is \(y=2\). The \(x\)-intercept is \(2\) (the value \(-2\) is the hole). The \(y\)-intercept is \(f(0)=\dfrac{-8}{-6}=\tfrac43\).

14 Sign chart for a quotient ★★★

Critical numbers: \(-4\) (included, numerator \(0\)) and \(2\) (excluded, denominator \(0\)). Signs on \((-\infty,-4)\), \((-4,2)\), \((2,\infty)\): \(\dfrac{-}{-}=+\), \(\dfrac{+}{-}=-\), \(\dfrac{+}{+}=+\). The solution is \((-\infty,-4]\cup(2,\infty)\).

15 Slant asymptote ★★★

The numerator has degree 2 and the denominator degree 1, so there is a slant asymptote. Divide with \(c=-2\):

So \(f(x)=2x-1-\dfrac{3}{x+2}\). The slant asymptote is \(y=2x-1\), and the vertical asymptote is \(x=-2\).

16 Build a polynomial ★★★

Since the coefficients are real, \(2+i\) is also a zero, and \((x-(2-i))(x-(2+i))=x^2-4x+5\). So \(P(x)=a(x+3)(x-1)^2(x^2-4x+5)\).

\(P(0)=a\cdot3\cdot1\cdot5=15a=30\), so \(a=2\). Thus \(P(x)=2(x+3)(x-1)^2(x^2-4x+5)\).

The degree is \(1+2+2=5\) and the leading coefficient is \(2\gt0\): the graph falls on the left and rises on the right.

17 From a graph to an equation ★★★

1. The graph touches the axis at \(-1\) (multiplicity 2) and crosses at \(2\) (multiplicity 1): \(f(x)=a(x+1)^2(x-2)\).
2. \(f(0)=a\cdot1\cdot(-2)=2\), so \(a=-1\).
3. \((x+1)^2(x-2)=(x^2+2x+1)(x-2)=x^3-3x-2\), so \(f(x)=-x^3+3x+2\). Check: \(f(2)=-8+6+2=0\). The negative leading coefficient also matches a graph that falls on the right.

18 Zero given, find the rest ★★★

The coefficients are real, so \(-i\) is also a zero, and \((x-i)(x+i)=x^2+1\) is a factor. Dividing, \(P(x)=(x^2+1)(x^2+x-6)\), since \((x^2+1)(x^2+x-6)=x^4+x^3-5x^2+x-6\). Then \(x^2+x-6=(x+3)(x-2)\). The zeros are \(i\), \(-i\), \(2\), and \(-3\).

19 A rational inequality with a twist ★★★

The sign of \(x-2\) is unknown, so multiplying could reverse the inequality. Instead subtract: \(\dfrac{x+1}{x-2}-2=\dfrac{x+1-2(x-2)}{x-2}=\dfrac{5-x}{x-2}\ge0\).

Critical numbers: \(5\) (included) and \(2\) (excluded). Signs on \((-\infty,2)\), \((2,5)\), \((5,\infty)\): \(\dfrac{+}{-}=-\), \(\dfrac{+}{+}=+\), \(\dfrac{-}{+}=-\). The solution is \((2,5]\). Check: at \(x=5\) the fraction is \(6/3=2\), and at \(x=3\) it is \(4\ge 2\).

20 A quartic inequality ★★★

Factor as a quadratic in \(x^2\): \(x^4-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)\). Critical numbers: \(-2,-1,1,2\). Signs on the five intervals: \(+,-,+,-,+\) (test \(x=-3\), \(-1.5\), \(0\), \(1.5\), \(3\)). We want the product \(\le0\), and the zeros are included. The solution is \([-2,-1]\cup[1,2]\).

21 Average cost ★★★

1. The degrees are equal, so the horizontal asymptote is \(y=\tfrac81=8\). As production grows, the average cost gets closer and closer to 8 dollars per item, the materials cost, as the fixed cost is spread over more items.
2. Solve \(\dfrac{8x+1200}{x}\le10\). Since \(x\gt0\), multiplying by \(x\) is safe: \(8x+1200\le10x\), so \(x\ge600\). Check: \(A(600)=\dfrac{6000}{600}=10\). At least 600 items are needed.

22 Missing term, fractional divisor ★★★

The \(x^2\) term is missing, so use coefficients \(2,0,-5,1\).

The quotient is \(2x^2+x-\tfrac92\) and the remainder is \(-\tfrac54\), so \(P\bigl(\tfrac12\bigr)=-\tfrac54\). Check directly: \(2\cdot\tfrac18-\tfrac52+1=\tfrac14-\tfrac52+1=-\tfrac54\). Because the remainder is not \(0\), \(x-\tfrac12\) is not a factor.