Polynomial Functions: practice solutions, Grade 11 – download the PDF

1 Is it a polynomial? ★★★

1. Yes. The exponents are whole numbers and \(\sqrt{2}\) is just a constant coefficient.
2. No. \(\dfrac{4}{x} = 4x^{-1}\) has a negative exponent.
3. No. \(\sqrt{x} = x^{1/2}\) has a fractional exponent.
4. Yes. \(j(x) = \dfrac{1}{3}x^2 + \dfrac{1}{3}\), a quadratic.

2 Standard form and degree ★★★

Standard form: \(f(x) = 7x^5 - 4x^3 - x + 9\).

The degree is \(5\), the leading coefficient is \(7\) and the constant term is \(9\).

3 Adding and subtracting ★★★

1. Combine like terms: \(3x^3 + 3x^2 - 2x - 6\).
2. Distribute the minus sign: \(6x^2 + x - 4 - 2x^2 + 3x - 9 = 4x^2 + 4x - 13\).

4 Multiplying binomials ★★★

(a) \(x^2 - 3x + 5x - 15 = x^2 + 2x - 15\).

(b) \(6x^2 + 8x - 3x - 4 = 6x^2 + 5x - 4\).

5 Evaluating with the Remainder Theorem ★★★

\(p(2) = 8 - 8 + 10 - 1 = 9\) and \(p(-1) = -1 - 2 - 5 - 1 = -9\).

By the Remainder Theorem, the remainder on division by \(x - 2\) is \(9\) and the remainder on division by \(x + 1\) is \(-9\).

6 A first synthetic division ★★★

With \(c = 2\) and coefficients \(1, 2, -5, -6\): bring down \(1\); \(2 + 2 = 4\); \(-5 + 8 = 3\); \(-6 + 6 = 0\).

The quotient is \(x^2 + 4x + 3\) and the remainder is \(0\). Since the remainder is zero, \(x - 2\) is a factor.

7 Cubes in two steps ★★★

(a) \(x^3 + 4^3 = (x + 4)(x^2 - 4x + 16)\).

(b) \((2x)^3 - 1^3 = (2x - 1)(4x^2 + 2x + 1)\).

8 Reading end behavior ★★★

Look at the degree and the sign of the leading coefficient.

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

9 Long division with a remainder ★★★

\(x^3 \div x = x^2\); \(x^2(x - 2) = x^3 - 2x^2\); subtract: \(-4x^2 + 11x\).

\(-4x^2 \div x = -4x\); \(-4x(x - 2) = -4x^2 + 8x\); subtract: \(3x - 7\).

\(3x \div x = 3\); \(3(x - 2) = 3x - 6\); subtract: \(-1\).

So \(x^3 - 6x^2 + 11x - 7 = (x - 2)(x^2 - 4x + 3) - 1\). The remainder is \(-1\), which agrees with \(p(2) = 8 - 24 + 22 - 7 = -1\).

10 Missing terms ★★★

Write the dividend as \(3x^3 + 0x^2 + 2x - 5\) and use \(c = -1\).

Bring down \(3\); \(0 - 3 = -3\); \(2 + 3 = 5\); \(-5 - 5 = -10\).

The quotient is \(3x^2 - 3x + 5\) and the remainder is \(-10\).

11 Synthetic division, degree 4 ★★★

The coefficients are \(2, -5, 0, -1, 6\) and \(c = 2\). Bring down \(2\); \(-5 + 4 = -1\); \(0 - 2 = -2\); \(-1 - 4 = -5\); \(6 - 10 = -4\).

The quotient is \(2x^3 - x^2 - 2x - 5\) and the remainder is \(-4\), so \(p(2) = -4\).

12 Finding a missing coefficient ★★★

By the Factor Theorem, \(p(-3) = 0\): \(-27 + 9k + 12 - 12 = 0\), so \(9k = 27\) and \(k = 3\).

Then \(p(x) = x^3 + 3x^2 - 4x - 12 = x^2(x + 3) - 4(x + 3) = (x + 3)(x^2 - 4) = (x + 3)(x - 2)(x + 2)\).

13 The packaging box ★★★

The volume is \(x(x + 4)(x - 1) = x^3 + 3x^2 - 4x\), so \(x^3 + 3x^2 - 4x - 42 = 0\).

Test \(x = 3\): \(27 + 27 - 12 - 42 = 0\). Synthetic division gives \(x^3 + 3x^2 - 4x - 42 = (x - 3)(x^2 + 6x + 14)\). The quadratic has discriminant \(36 - 56 < 0\), so \(x = 3\) is the only real solution.

The box is \(3\) in by \(7\) in by \(2\) in, which is about \(7.6\) cm by \(17.8\) cm by \(5.1\) cm. Check: \(3 \times 7 \times 2 = 42\).

14 Rational root candidates ★★★

Divisors of \(10\): \(\pm 1, \pm 2, \pm 5, \pm 10\). Divisors of \(3\): \(\pm 1, \pm 3\). Candidates: \(\pm 1, \pm 2, \pm 5, \pm 10, \pm\dfrac{1}{3}, \pm\dfrac{2}{3}, \pm\dfrac{5}{3}, \pm\dfrac{10}{3}\).

\(p(1) = 3 - 1 + 8 - 10 = 0\). Synthetic division gives \(3x^2 + 2x + 10\).

The discriminant is \(4 - 120 = -116 < 0\), so \(x = \dfrac{-2 \pm 2i\sqrt{29}}{6} = \dfrac{-1 \pm i\sqrt{29}}{3}\). The only real zero is \(1\); the other two are non-real conjugates.

15 Sum and difference of cubes ★★★

(a) \((5x)^3 + 3^3 = (5x + 3)(25x^2 - 15x + 9)\).

(b) First factor out \(2\): \(54x^3 - 2 = 2(27x^3 - 1) = 2(3x - 1)(9x^2 + 3x + 1)\).

16 Reading a graph ★★★

1. The graph crosses the x-axis at \(-2\), \(1\) and \(3\).
2. With leading coefficient \(1\): \(k(x) = (x + 2)(x - 1)(x - 3)\). Check: \(k(0) = 2 \times (-1) \times (-3) = 6\). Expanded, \(k(x) = x^3 - 2x^2 - 5x + 6\).
3. The degree is odd and the leading coefficient is positive, so the graph falls on the left and rises on the right.

17 Solving a cubic completely ★★★

Candidates: \(\pm 1, \pm 2, \pm 3, \pm 6, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}\). Try \(x = -2\): \(-16 - 4 + 26 - 6 = 0\).

Synthetic division with \(c = -2\): \(2\); \(-1 - 4 = -5\); \(-13 + 10 = -3\); \(-6 + 6 = 0\). The quotient is \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\).

So \((x + 2)(2x + 1)(x - 3) = 0\) and the solutions are \(-2\), \(-\dfrac{1}{2}\) and \(3\).

18 Building a polynomial from its zeros ★★★

Real coefficients force the conjugate \(1 - 3i\) to be a zero too. Then \((x - (1 + 3i))(x - (1 - 3i)) = (x - 1)^2 - (3i)^2 = x^2 - 2x + 10\).

Multiply by \((x - 2)\): \((x - 2)(x^2 - 2x + 10) = x^3 - 2x^2 + 10x - 2x^2 + 4x - 20 = x^3 - 4x^2 + 14x - 20\).

The degree is \(3\), matching the three zeros \(2\), \(1 + 3i\), \(1 - 3i\).

19 Six zeros ★★★

\(x^6 - 64 = (x^2)^3 - 4^3 = (x^2 - 4)(x^4 + 4x^2 + 16)\); instead split as \((x^3 - 8)(x^3 + 8)\).

\(x^3 - 8 = (x - 2)(x^2 + 2x + 4)\) and \(x^3 + 8 = (x + 2)(x^2 - 2x + 4)\).

The linear factors give \(2\) and \(-2\). The quadratics give \(x = \dfrac{-2 \pm \sqrt{4 - 16}}{2} = -1 \pm i\sqrt{3}\) and \(x = 1 \pm i\sqrt{3}\).

Six zeros for degree \(6\): \(\pm 2,\ -1 \pm i\sqrt{3},\ 1 \pm i\sqrt{3}\).

20 Two conditions, two unknowns ★★★

\(p(1) = 1 + a + b - 6 = -8\) gives \(a + b = -3\). \(p(-1) = -1 + a - b - 6 = 0\) gives \(a - b = 7\).

Adding the equations: \(2a = 4\), so \(a = 2\) and \(b = -5\).

Then \(p(x) = x^3 + 2x^2 - 5x - 6 = (x + 1)(x^2 + x - 6) = (x + 1)(x + 3)(x - 2)\).

21 True or false? ★★★

1. True. The two ends go in opposite directions, so the graph must cross the x-axis somewhere.
2. True. For example, \(x^4 + 1\) is never \(0\) for real \(x\); its four zeros are all non-real.
3. False. The theorem counts \(5\) zeros with multiplicity, in the complex numbers. For instance \(x^5\) has the single real zero \(0\) with multiplicity \(5\).