Polynomial, Rational, and Radical Relationships
Unit 1 of Algebra II. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Complex numbers, polynomial identities and arithmetic, the remainder theorem, zeros and graphs of polynomials, rational expressions, radical and rational equations, systems.
How this unit is tested
Start by sorting problems into two families: those about the algebra of numbers and expressions (complex numbers, polynomial identities, rational expressions) and those about the behavior of functions (zeros, graphs, equations, systems). For the first family, the work is mechanical but exacting — track signs, remember $i^2=-1$, and always state domain restrictions when you cancel a factor from a rational expression.
For the second family, connect algebra to graphs at every step. A zero found by synthetic division or the factor theorem is an x-intercept; its multiplicity tells you whether the graph crosses or bounces there; the degree and leading coefficient tell you what happens at the far left and right. When a real polynomial has a non-real zero, its conjugate is automatically a zero too — use that to cut a cubic down to a quadratic you can solve directly.
For equations with radicals or rational expressions, the danger is not the algebra but the aftermath: squaring a radical equation or multiplying a rational equation by a variable expression can introduce solutions that do not satisfy the original equation. Always substitute your solutions back into the equation as it was originally written, not the cleared-out version.
For systems, decide first whether you are intersecting a line with a curve (substitution is fastest) or two curves; solve for one variable, substitute back, and verify every point in both original equations before reporting it as an answer.
What you have to know
Remainder Theorem
If a polynomial p(x) is divided by (x-c), the remainder equals p(c).
Factor Theorem
(x-c) is a factor of p(x) if and only if p(c)=0.
Fundamental Theorem of Algebra
A polynomial of degree n has exactly n complex zeros, counting multiplicity.
Complex Conjugate Root Theorem
If a polynomial has real coefficients and a+bi (with b not 0) is a zero, then its conjugate a-bi is also a zero.
Sum and Difference of Cubes
$a^3+b^3=(a+b)(a^2-ab+b^2)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$.
Rational Root Theorem
For a polynomial with integer coefficients, every rational zero p/q (in lowest terms) has p dividing the constant term and q dividing the leading coefficient.
14 practice questions
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Simplify $(3+2i)(1-4i)$ and write the result in the form a+bi.
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Answer. 11-10i
Distribute: $3(1)+3(-4i)+2i(1)+2i(-4i)=3-12i+2i-8i^2$. Since $i^2=-1$, $-8i^2=8$, giving $11-10i$. -
Solve $x^2+4x+13=0$ over the complex numbers.
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Answer. x = -2+3i or x = -2-3i
By the quadratic formula, the discriminant is $16-52=-36$, so $\sqrt{-36}=6i$. Then $x=\dfrac{-4\pm6i}{2}=-2\pm3i$. -
Use synthetic division to divide $x^3-4x^2+x+6$ by $x-3$, and use the Remainder Theorem to state the remainder.
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Answer. Quotient x^2-x-2, remainder 0
Synthetic division with c=3 on coefficients 1,-4,1,6 gives 1, -1, -2, and a final 0. The remainder 0 equals p(3), confirming x-3 is a factor. -
Determine whether (x+1) is a factor of $p(x)=x^3+2x^2-5x-6$, using the Factor Theorem.
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Answer. Yes, because p(-1)=0
Evaluate $p(-1)=-1+2+5-6=0$. Since the remainder of dividing by (x+1) is 0, the Factor Theorem confirms (x+1) is a factor. -
One zero of $q(x)=x^3-4x^2+9x-36$ is $3i$. Find the other two zeros.
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Answer. -3i and 4
By the Complex Conjugate Root Theorem, $-3i$ is also a zero. Multiplying $(x-3i)(x+3i)=x^2+9$ and dividing q(x) by $x^2+9$ leaves quotient $x-4$, giving the third zero 4. -
Describe the behavior of the graph of $f(x)=(x-1)^2(x+3)$ at each of its zeros.
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Answer. At x=1 the graph touches the x-axis and turns back (even multiplicity 2); at x=-3 it crosses the x-axis (multiplicity 1).
Even multiplicity produces a tangent point; odd multiplicity produces a crossing, matching the factors' exponents. -
Describe the end behavior of $f(x)=-2x^4+3x^2-1$.
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Answer. As x approaches positive or negative infinity, f(x) approaches negative infinity on both sides.
The degree is even (4) and the leading coefficient is negative, so both ends of the graph point downward. -
Simplify $\dfrac{x^2-9}{x^2+x-12}$ and state any excluded values of x.
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Answer. $\dfrac{x+3}{x+4}$, with x not equal to 3 or -4
Factor to get $\dfrac{(x-3)(x+3)}{(x+4)(x-3)}$, cancel (x-3) noting x cannot equal 3, and the remaining denominator forbids x=-4. -
Add and simplify $\dfrac{2}{x-1}+\dfrac{3}{x+2}$.
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Answer. $\dfrac{5x+1}{(x-1)(x+2)}$
Using common denominator (x-1)(x+2), the numerator becomes $2(x+2)+3(x-1)=2x+4+3x-3=5x+1$. -
Solve $\dfrac{x}{x-2}+1=\dfrac{2}{x-2}$, checking for extraneous solutions.
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Answer. No solution (x=2 is extraneous)
Multiplying by (x-2) gives $x+(x-2)=2$, so $2x-2=2$ and $x=2$. But x=2 makes the original denominators zero, so it must be rejected, leaving no valid solution. -
Solve $\sqrt{x+7}=x+1$, checking for extraneous roots.
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Answer. x=2
Squaring gives $x+7=x^2+2x+1$, so $x^2+x-6=0$, factoring to $(x+3)(x-2)=0$. Testing x=-3 in the original equation fails (2 ≠ -2), so only x=2 works. -
Solve the system $y=x^2-x-6$ and $y=x+2$.
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Answer. (4,6) and (-2,0)
Set $x^2-x-6=x+2$, giving $x^2-2x-8=0$, which factors as $(x-4)(x+2)=0$. Substituting x=4 and x=-2 into y=x+2 gives y=6 and y=0, and both points check in the original quadratic. -
Factor $8x^3+27$ using the sum of cubes identity.
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Answer. $(2x+3)(4x^2-6x+9)$
With $a=2x$ and $b=3$, $a^3+b^3=(a+b)(a^2-ab+b^2)$ gives $(2x+3)((2x)^2-(2x)(3)+3^2)=(2x+3)(4x^2-6x+9)$. -
Use the Binomial Theorem to find the coefficient of $x^2$ in the expansion of $(x-2)^4$.
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Answer. 24
The $x^2$ term is $\binom{4}{2}x^2(-2)^2=6\cdot4\cdot x^2=24x^2$, so the coefficient is 24.
What people get wrong
- Squaring both sides of a radical equation and stopping there — this can introduce extraneous roots. Always substitute each candidate solution back into the original (unsquared) equation.
- Treating i like an ordinary variable when multiplying complex numbers and forgetting that $i^2=-1$. Simplify every $i^2$ term you produce before combining like terms.
- Setting up synthetic division with the wrong sign — dividing by $(x-c)$ uses c, but dividing by $(x+c)$ requires using $-c$. Rewrite the divisor in $(x-c)$ form first.
- Canceling a common factor in a rational expression without noting that the canceled factor's zero is still excluded from the domain. State the restriction before or as you simplify.
- Assuming every zero makes the graph cross the x-axis. A zero with even multiplicity only touches the axis and turns back; only odd multiplicity produces a crossing.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Algebra II. Free, and no account needed to start.