Complex Factorization and Volume of a Sphere: Precalculus Math Worksheet Answer Key – PDF Download

Precalculus Complex Identities Worksheet Answer Key: problems 1–5

Problem 1: Powers of i

The powers of \(i\) repeat every 4 steps: \(i, -1, -i, 1\). Divide the exponent by 4 and use the remainder.

1. \(5=4+1\), so \(i^{5}=i\). \(i\)
2. \(10=8+2\), so \(i^{10}=i^{2}=-1\). \(-1\)
3. \(27=24+3\), so \(i^{27}=i^{3}=-i\). \(-i\)
4. \(100=4(25)\), so \(i^{100}=1\). \(1\)

Problem 2: Expand a conjugate product

\((x+3i)(x-3i)=x^{2}-3ix+3ix-9i^{2}=x^{2}-9(-1)=x^{2}+9\). \(x^{2}+9\)

Problem 3: Factor sums of squares

1. \(x^{2}+5^{2}=(x+5i)(x-5i)\). \((x+5i)(x-5i)\)
2. \(x^{2}+7^{2}=(x+7i)(x-7i)\). \((x+7i)(x-7i)\)
3. \(y^{2}+1^{2}=(y+i)(y-i)\). \((y+i)(y-i)\)

Problem 4: Leading coefficients

1. \((2x)^{2}+1^{2}=(2x+i)(2x-i)\). \((2x+i)(2x-i)\)
2. \((3x)^{2}+4^{2}=(3x+4i)(3x-4i)\). \((3x+4i)(3x-4i)\)
3. \((7y)^{2}+2^{2}=(7y+2i)(7y-2i)\). \((7y+2i)(7y-2i)\)

Problem 5: Irrational constant

\(x^{2}+7=x^{2}+(\sqrt{7})^{2}=(x+i\sqrt{7})(x-i\sqrt{7})\). \((x+i\sqrt{7})(x-i\sqrt{7})\)

Precalculus Complex Identities Worksheet Answer Key: problems 6–10

Problem 6: Factor out a common factor

1. \(2(x^{2}+9)=2(x+3i)(x-3i)\). \(2(x+3i)(x-3i)\)
2. \(5(x^{2}+9)=5(x+3i)(x-3i)\). \(5(x+3i)(x-3i)\)
3. \(3(x^{2}+25)=3(x+5i)(x-5i)\). \(3(x+5i)(x-5i)\)

Problem 7: Solve over the complex numbers

1. \(x^{2}=-36\), so \(x=\pm 6i\). \(x=\pm 6i\)
2. \(x^{2}=-12\), so \(x=\pm i\sqrt{12}=\pm 2i\sqrt{3}\). \(x=\pm 2i\sqrt{3}\)
3. \(x^{2}=-\frac{9}{4}\), so \(x=\pm\frac{3}{2}i\). \(x=\pm\frac{3}{2}i\)

Problem 8: Write a polynomial from its zeros

The factors are \((x-3i)\) and \((x+3i)\). \((x-3i)(x+3i)=x^{2}-9i^{2}=x^{2}+9\). \(x^{2}+9\)

Problem 9: Fourth-degree difference

\(x^{4}-16=(x^{2}-4)(x^{2}+4)\). Then \(x^{2}-4=(x-2)(x+2)\) and \(x^{2}+4=(x+2i)(x-2i)\). \((x-2)(x+2)(x+2i)(x-2i)\)

Problem 10: Another fourth-degree difference

\(x^{4}-81=(x^{2}-9)(x^{2}+9)=(x-3)(x+3)(x+3i)(x-3i)\). \((x-3)(x+3)(x+3i)(x-3i)\)

Precalculus Complex Identities Worksheet Answer Key: problems 11–15

Problem 11: Cubic with a common factor

1. \(x(x^{2}+4)=x(x+2i)(x-2i)\). \(x(x+2i)(x-2i)\)
2. \(x(x^{2}+1)=x(x+i)(x-i)\). \(x(x+i)(x-i)\)

Problem 12: Factor by grouping

Group: \(x^{2}(x-2)+4(x-2)=(x-2)(x^{2}+4)=(x-2)(x+2i)(x-2i)\). \((x-2)(x+2i)(x-2i)\), with zeros \(2, 2i, -2i\).

Problem 13: Products of conjugates

1. \(9-4i^{2}=9+4=13\). \(13\)
2. \(25-i^{2}=26\). \(26\)
3. \(36-49i^{2}=85\). \(85\)

Each product is \(a^{2}+b^{2}\), a real number.

Problem 14: Numbers as conjugate products

1. \(41=5^{2}+4^{2}=(5+4i)(5-4i)\). \((5+4i)(5-4i)\)
2. \(85=9^{2}+2^{2}=(9+2i)(9-2i)\). \((9+2i)(9-2i)\)
3. \(65=8^{2}+1^{2}=(8+i)(8-i)\). \((8+i)(8-i)\)

Check: \((5+4i)(5-4i)=25+16=41\), and the others work the same way.

Problem 15: Evaluate at complex values

\(p(2i)=(2i)^{2}+4=-4+4=0\). \(p(-2i)=(-2i)^{2}+4=-4+4=0\). \(p(1+i)=(1+2i+i^{2})+4=(1+2i-1)+4=4+2i\).

\(0, 0, 4+2i\). The first two values confirm that \(2i\) and \(-2i\) are zeros.

Precalculus Complex Identities Worksheet Answer Key: problems 16–20

Problem 16: Identities with complex numbers

1. \((z+2i)^{2}=z^{2}+2z(2i)+(2i)^{2}=z^{2}+4iz-4\). \(z^{2}+4iz-4\)
2. \(4+4i+i^{2}=3+4i\). \(3+4i\)

Problem 17: Complete the square, then factor

\(x^{2}-2x+5=(x^{2}-2x+1)+4=(x-1)^{2}+2^{2}=(x-1+2i)(x-1-2i)\). \((x-1+2i)(x-1-2i)\), with zeros \(1\pm 2i\).

Problem 18: Another quadratic

\(x^{2}+6x+13=(x^{2}+6x+9)+4=(x+3)^{2}+2^{2}=(x+3+2i)(x+3-2i)\). \((x+3+2i)(x+3-2i)\), zeros \(-3\pm 2i\).

Problem 19: Volume of a sphere from radius or diameter

1. \(V=\frac{4}{3}\pi(3)^{3}=\frac{4}{3}\pi(27)=36\pi\approx 113.10\) cubic cm. \(36\pi\)
2. The radius is \(5\). \(V=\frac{4}{3}\pi(125)=\frac{500\pi}{3}\approx 523.60\) cubic in. \(\frac{500\pi}{3}\)

Problem 20: Volume of a hemisphere

\(V=\frac{2}{3}\pi(6)^{3}=\frac{2}{3}\pi(216)=144\pi\). \(144\pi\) cubic feet

Precalculus Complex Identities Worksheet Answer Key: problems 21–25

Problem 21: Cylinder minus cone

Cylinder: \(\pi(16)(4)=64\pi\). Cone: \(\frac{1}{3}\pi(16)(4)=\frac{64\pi}{3}\). Difference: \(64\pi-\frac{64\pi}{3}=\frac{128\pi}{3}\). Hemisphere: \(\frac{2}{3}\pi(64)=\frac{128\pi}{3}\). Both equal \(\frac{128\pi}{3}\) cubic inches.

Problem 22: Matching cross-sections

1. The radius is \(\sqrt{5^{2}-3^{2}}=\sqrt{16}=4\), so the area is \(16\pi\) square cm.
2. The ring has outer radius \(5\) and inner radius \(3\), so the area is \(25\pi-9\pi=16\pi\) square cm.

Both areas are \(16\pi\), as Cavalieri’s principle requires.

Problem 23: Leaning stack of coins

1. Height: \(30(0.1)=3\) in. \(V=\pi(0.5)^{2}(3)=0.75\pi\approx 2.36\) cubic in. \(0.75\pi\)
2. The same, \(0.75\pi\) cubic in. Every level cuts both stacks in a circle of area \(0.25\pi\) square in., so by Cavalieri’s principle the volumes are equal.

Problem 24: Melted ice cream

Sphere: \(\frac{4}{3}\pi(1.5)^{3}=\frac{4}{3}\pi(3.375)=4.5\pi\). Cup: \(\pi(1.5)^{2}h=2.25\pi h\). Set equal: \(2.25\pi h=4.5\pi\), so \(h=2\). 2 inches deep

Problem 25: Tennis ball can

The can height is \(3(3)=9\) in. Can volume: \(\pi(2.25)(9)=20.25\pi\). One ball: \(\frac{4}{3}\pi(3.375)=4.5\pi\), so three balls: \(13.5\pi\). Empty space: \(20.25\pi-13.5\pi=6.75\pi\approx 21.21\). About 21.21 cubic inches

Precalculus Complex Identities Worksheet Answer Key: problems 26–28

Problem 26: Spherical water tank

\(V=\frac{4}{3}\pi(12)^{3}=\frac{4}{3}\pi(1728)=2304\pi\approx 7238.2\) cubic feet. Gallons: \(2304\pi \times 7.48\approx 54142\). About 7,238.2 cubic feet, or about 54,142 gallons

Problem 27: Basketball volume

The radius is \(4.75\). \(V=\frac{4}{3}\pi(4.75)^{3}=\frac{4}{3}\pi(107.171875)\approx 448.9\). About 448.9 cubic inches

Problem 28: Sphere and cylinder ratio

Cylinder: \(\pi r^{2}(2r)=2\pi r^{3}\). Sphere: \(\frac{4}{3}\pi r^{3}\). Ratio: \(\frac{4}{3}: 2=\frac{2}{3}\). For \(r=3\): cylinder \(\pi(9)(6)=54\pi\), sphere \(36\pi\), and \(\frac{36}{54}=\frac{2}{3}\). The ratio is \(\frac{2}{3}\)

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