Algebra 2 Zeros and Graphs of Polynomial Functions Worksheet Solutions (PDF)

Set each factor equal to 0: \(x – 3 = 0\), \(x + 5 = 0\), \(2x – 1 = 0\).

\(x = 3, -5, \frac{1}{2}\)

\(x^{3} – 9x = x(x^{2} – 9) = x(x – 3)(x + 3)\).

\(x = 0, 3, -3\)

\(x(x^{2} + 2x – 8) = x(x + 4)(x – 2)\).

\(x = 0, -4, 2\)

Group: \(x^{2}(x + 3) – 4(x + 3) = (x + 3)(x^{2} – 4)\).

\((x + 3)(x – 2)(x + 2)\).

\(x = -3, 2, -2\)

Let \(u = x^{2}\): \(u^{2} – 5u + 4 = (u – 1)(u – 4)\).

So \(x^{2} = 1\) or \(x^{2} = 4\).

\(x = \pm 1, \pm 2\)

\(x = -1\) has multiplicity 2 (even): the graph touches the axis.

\(x = 2\) has multiplicity 3 (odd): the graph crosses the axis.

The degree is \(2 + 3 = 5\).

\(-1\): multiplicity 2, touches; \(2\): multiplicity 3, crosses

The graph crosses at \(-2\) (multiplicity 1) and touches at \(1\) (multiplicity 2). The total degree is 3.

\(f(x) = (x + 2)(x – 1)^{2}\)

The leading term is \(-2x^{5}\): odd degree, negative coefficient.

As x decreases without bound, f(x) increases without bound; as x increases without bound, f(x) decreases without bound

The leading term \(3x^{4}\) has even degree and a positive coefficient.

Both ends rise: f(x) increases without bound as x goes to either end

Factors of the constant term 6: \(\pm 1, \pm 2, \pm 3, \pm 6\). Factors of the leading coefficient 2: \(\pm 1, \pm 2\).

Divide each by 1 and 2.

\(\pm 1, \pm 2, \pm 3, \pm 6, \pm\frac{1}{2}, \pm\frac{3}{2}\)

Candidates: \(\pm 1, \pm 2, \pm 3, \pm 6\). \(p(1) = 1 – 2 – 5 + 6 = 0\), so \(x – 1\) is a factor.

Divide: \(x^{3} – 2x^{2} – 5x + 6 = (x – 1)(x^{2} – x – 6) = (x – 1)(x – 3)(x + 2)\).

\(x = 1, 3, -2\)

\(p(1) = -6\), no. \(p(3) = 54 – 27 – 33 + 6 = 0\), so \(x – 3\) is a factor.

Synthetic division gives \(2x^{2} + 3x – 2 = (2x – 1)(x + 2)\).

\(x = 3, \frac{1}{2}, -2\)

\(f(x) = (x – 2)(x + 1)(x – 4)\).

\((x – 2)(x + 1) = x^{2} – x – 2\). Multiply by \(x – 4\).

\(x^{3} – 5x^{2} + 2x + 8\)

\(f(x) = a(x – 1)^{2}(x + 3)\). At \(x = 0\): \(a(1)(3) = 6\), so \(a = 2\).

Expand \(2(x^{2} – 2x + 1)(x + 3)\).

\(2x^{3} + 2x^{2} – 10x + 6\)

Test points: \(f(-2) = (-1)(-4)(-6) = -24\), \(f(0) = (1)(-2)(-4) = 8\), \(f(3) = (4)(1)(-1) = -4\), \(f(5) = (6)(3)(1) = 18\).

negative, positive, negative, positive

\(x^{3} – 1 = (x – 1)(x^{2} + x + 1)\). The first factor gives \(x = 1\).

Quadratic formula: \(x = \frac{-1 \pm \sqrt{1 – 4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}\).

\(x = 1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 – i\sqrt{3}}{2}\)

Complex zeros of a real polynomial come in conjugate pairs, so \(-3i\) is also a zero.

Count: \(2 + 1 + 1 + 1 = 5\), so there are no others.

\(-3i\)

\(V(x) = x(12 – 2x)(8 – 2x)\), so the zeros are \(x = 0, 6, 4\).

A cut cannot exceed half of the shorter side (4 inches), so only \(0 < x < 4\) makes a box.

The maximum is at \(x \approx 1.57\) inches, where \(V \approx 67.6\) cubic inches.

Zeros 0, 4, 6; domain 0 < x < 4; best cut about 1.57 in

Zeros: \(x = 0\) (multiplicity 1) and \(x = 4\) (multiplicity 2).

At \(x = 4\) the graph touches the ground and turns back up. \(h(2) = 2(-2)^{2} = 8\).

Ground at x = 0 and x = 4 (touches at 4); h(2) = 8 tens of feet = 80 ft

The factor \(x – 2\) is squared, so \(x = 2\) has even multiplicity 2. The graph touches the axis and turns around, it does not cross.

Maya is wrong; the graph touches at x = 2 and crosses at x = -1

The ends of an odd-degree graph go in opposite directions: one end goes to positive infinity and the other to negative infinity.

A continuous graph that goes from below the x-axis to above it must cross the axis at least once.

At least one x-intercept, hence one real zero