This answer key gives a full worked solution for each of the 22 polynomial division problems, including every subtraction in the long division and every row of the synthetic division tables. Check your work against the final bold answer.
Problem 1: Divide by a monomial
Divide each term by the monomial.
a) \(\frac{12x^{3}}{4x}=3x^{2}\), \(\frac{-8x^{2}}{4x}=-2x\), \(\frac{20x}{4x}=5\).
b) \(\frac{9x^{4}}{3x^{2}}=3x^{2}\), \(\frac{6x^{3}}{3x^{2}}=2x\), \(\frac{-15x^{2}}{3x^{2}}=-5\).
a) \(3x^{2}-2x+5\) b) \(3x^{2}+2x-5\)
Problem 2: Missing side of a rectangle
Area equals length times width, so divide the area by \(x+3\).
\(x^{2}: x=x\). Multiply: \(x(x+3)=x^{2}+3x\). Subtract: \(4x+12\). Then \(4x: x=4\), and \(4(x+3)=4x+12\), so the remainder is \(0\).
The other side is \(x+4\).
Problem 3: Long division, zero remainder
a) \(x^{2}: x=x\); subtract \(x^{2}+2x\) to get \(-7x-14\); \(-7x: x=-7\); subtract \(-7x-14\) to get \(0\).
b) \(x^{2}: x=x\); subtract \(x^{2}+4x\) to get \(7x+28\); \(7x: x=7\); subtract \(7x+28\) to get \(0\).
a) \(x-7\) b) \(x+7\) (both remainders are \(0\))
Problem 4: Long division with a remainder
a) \(x^{2}: x=x\); subtract \(x^{2}-x\) to get \(4x-2\); \(4x: x=4\); subtract \(4x-4\) to get \(2\).
b) \(x^{2}: x=x\); subtract \(x^{2}-3x\) to get \(-3x+4\); \(-3x: x=-3\); subtract \(-3x+9\) to get \(-5\).
a) \(x+4+\frac{2}{x-1}\) b) \(x-3-\frac{5}{x-3}\)
Problem 5: Cubic by long division
\(x^{3}: x=x^{2}\); subtract \(x^{3}-2x^{2}\) to get \(6x^{2}-3x\).
\(6x^{2}: x=6x\); subtract \(6x^{2}-12x\) to get \(9x-18\).
\(9x: x=9\); subtract \(9x-18\) to get \(0\).
Quotient \(x^{2}+6x+9\), remainder \(0\).
Problem 6: Division with missing terms
a) Write \(x^{3}+0x^{2}+0x-8\). The quotient terms are \(x^{2}\), \(2x\), and \(4\), with products \(x^{3}-2x^{2}\), \(2x^{2}-4x\), and \(4x-8\). The remainder is \(0\).
b) Write \(x^{3}+0x^{2}+0x+27\). The quotient terms are \(x^{2}\), \(-3x\), and \(9\), with products \(x^{3}+3x^{2}\), \(-3x^{2}-9x\), and \(9x+27\). The remainder is \(0\).
a) \(x^{2}+2x+4\) b) \(x^{2}-3x+9\)
Problem 7: Fill in a synthetic division table
Bring down \(1\). Multiply \(1\cdot1=1\) and add to \(-6\) to get \(-5\). Multiply \(-5\cdot1=-5\) and add to \(11\) to get \(6\). Multiply \(6\cdot1=6\) and add to \(-6\) to get \(0\).
Quotient \(x^{2}-5x+6\), remainder \(0\).
Problem 8: Synthetic division with a remainder
Use \(c=3\) with coefficients \(2,\ -5,\ 1,\ 7\). Bring down \(2\). Then \(-5+6=1\), \(1+3=4\), and \(7+12=19\).
Quotient \(2x^{2}+x+4\), remainder \(19\), so the result is \(2x^{2}+x+4+\frac{19}{x-3}\).
Problem 9: Divide by x plus a number
a) \(c=-2\): \(1\); \(5-2=3\); \(-2-6=-8\); \(-24+16=-8\).
b) \(c=-1\) with coefficients \(1,\ 0,\ -3,\ 1,\ 5\): \(1\); \(0-1=-1\); \(-3+1=-2\); \(1+2=3\); \(5-3=2\).
a) \(x^{2}+3x-8\), remainder \(-8\) b) \(x^{3}-x^{2}-2x+3\), remainder \(2\)
Problem 10: Remainder Theorem
a) \(p(3)=27-36+6+5=2\).
b) \(p(-1)=-1-4-2+5=-2\).
c) \(p(2)=8-16+4+5=1\).
a) \(2\) b) \(-2\) c) \(1\)
Problem 11: Use the Factor Theorem
a) \(f(2)=8-12-20+24=0\), so \(x-2\) is a factor.
b) \(f(3)=27-27-30+24=-6\), which is not \(0\), so \(x-3\) is not a factor.
c) \(f(-3)=-27-27+30+24=0\), so \(x+3\) is a factor.
a) yes b) no c) yes
Problem 12: Find the missing coefficient
By the Factor Theorem, the value at \(x=2\) must be \(0\): \(8+4k-8-12=0\), so \(4k-12=0\) and \(k=3\).
Check: \(x^{3}+3x^{2}-4x-12=(x+3)(x^{2}-4)\), which has the factor \(x-2\).
\(k=3\)
Problem 13: Divide by a binomial with leading coefficient 2
a) \(6x^{2}: 2x=3x\); subtract \(6x^{2}+9x\) to get \(-8x-12\); \(-8x: 2x=-4\); subtract \(-8x-12\) to get \(0\).
b) \(4x^{2}: 2x=2x\); subtract \(4x^{2}-2x\) to get \(-2x+5\); \(-2x: 2x=-1\); subtract \(-2x+1\) to get \(4\).
a) \(3x-4\) b) \(2x-1+\frac{4}{2x-1}\)
Problem 14: Write the division statement
Use \(c=-1\): \(3\); \(2-3=-1\); \(-1+1=0\); \(4-0=4\). The quotient is \(3x^{2}-x+0\) and the remainder is \(4\).
Check: \((x+1)(3x^{2}-x)+4=3x^{3}-x^{2}+3x^{2}-x+4=3x^{3}+2x^{2}-x+4\).
\(p(x)=(x+1)(3x^{2}-x)+4\)
Problem 15: Factor a cubic from its graph
Divide by \(x-1\) with \(c=1\): \(1\); \(-2+1=-1\); \(-5-1=-6\); \(6-6=0\). The quotient is \(x^{2}-x-6\).
Factor: \(x^{2}-x-6=(x-3)(x+2)\).
\(f(x)=(x-1)(x-3)(x+2)\); zeros \(-2\), \(1\), \(3\).
Problem 16: Find the remaining zeros
Divide by \(x+1\) with \(c=-1\): \(1\); \(-2-1=-3\); \(-7+3=-4\); \(8+4=12\); \(12-12=0\). The quotient is \(x^{3}-3x^{2}-4x+12\).
Divide that by \(x-2\) with \(c=2\): \(1\); \(-3+2=-1\); \(-4-2=-6\); \(12-12=0\). The quotient is \(x^{2}-x-6=(x-3)(x+2)\).
The other zeros are \(x=3\) and \(x=-2\). (So \(g(x)=(x+1)(x-2)(x-3)(x+2)\).)
Problem 17: Verify a polynomial identity
Left side: \((x^{2}+y^{2})^{2}=x^{4}+2x^{2}y^{2}+y^{4}\).
Right side: \((x^{2}-y^{2})^{2}=x^{4}-2x^{2}y^{2}+y^{4}\) and \((2xy)^{2}=4x^{2}y^{2}\). Add: \(x^{4}+2x^{2}y^{2}+y^{4}\).
Both sides equal \(x^{4}+2x^{2}y^{2}+y^{4}\), so the identity holds.
Problem 18: Pythagorean triples from the identity
a) \(2xy=2\cdot3\cdot2=12\); \(x^{2}-y^{2}=9-4=5\); \(x^{2}+y^{2}=9+4=13\).
b) \(5^{2}+12^{2}=25+144=169=13^{2}\).
c) \(2xy=20\); \(x^{2}-y^{2}=25-4=21\); \(x^{2}+y^{2}=25+4=29\). Check: \(21^{2}+20^{2}=441+400=841=29^{2}\).
a) \(12,\ 5,\ 13\) b) \(169=169\) c) \(20,\ 21,\ 29\)
Problem 19: Height of a storage box
a) Height \(=V: (\text{length}\cdot\text{width})\). The product \((x+4)(x+2)=x^{2}+6x+8\). Divide: \(x^{3}: x^{2}=x\); subtract \(x^{3}+6x^{2}+8x\) to get \(3x^{2}+18x+24\); \(3x^{2}: x^{2}=3\); subtract \(3x^{2}+18x+24\) to get \(0\).
b) For \(x=3\): length \(7\) ft, width \(5\) ft, height \(6\) ft. Volume \(7\cdot5\cdot6=210\) ft\(^{3}\). Check: \(V(3)=27+81+78+24=210\).
a) \(x+3\) feet b) \(7\times 5\times 6\) ft, \(210\) cubic feet
Problem 20: Break-even points
Divide by \(x-2\) with \(c=2\): \(1\); \(-10+2=-8\); \(31-16=15\); \(-30+30=0\). The quotient is \(x^{2}-8x+15=(x-3)(x-5)\).
The other break-even points are \(x=3\) and \(x=5\), which means \(300\) and \(500\) shirts.
Problem 21: Find two unknown coefficients
Remainder Theorem: \(p(1)=1+a+b-6=0\), so \(a+b=5\). Also \(p(-1)=-1+a-b-6=-10\), so \(a-b=-3\).
Add the equations: \(2a=2\), so \(a=1\). Then \(b=4\).
Check: \(p(x)=x^{3}+x^{2}+4x-6\); \(p(1)=1+1+4-6=0\) and \(p(-1)=-1+1-4-6=-10\).
\(a=1\), \(b=4\)
Problem 22: Garden dimensions
a) Divide by \(2x+1\): \(2x^{3}: 2x=x^{2}\); subtract \(2x^{3}+x^{2}\) to get \(6x^{2}-5x\); \(6x^{2}: 2x=3x\); subtract \(6x^{2}+3x\) to get \(-8x-4\); \(-8x: 2x=-4\); subtract \(-8x-4\) to get \(0\).
b) \(x^{2}+3x-4=(x+4)(x-1)\).
c) For \(x=3\): width \(7\) ft, length \(9+9-4=14\) ft, area \(7\cdot14=98\) ft\(^{2}\). Check: \(A(3)=54+63-15-4=98\).
a) \(x^{2}+3x-4\) ft b) \((x+4)(x-1)\) c) \(14\) ft, \(7\) ft, \(98\) square feet
See every chapter on the Algebra 2 math page.