Expressions and Equations
Unit 4 of Algebra I. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Polynomial arithmetic, factoring, completing the square, the quadratic formula, solving quadratic equations, systems with one linear and one quadratic equation.
How this unit is tested
Start by sorting any given problem into one of two buckets: is it an expression to simplify, or an equation to solve? For expressions, apply the distributive property carefully (line up like terms by degree) before combining anything. For equations that look quadratic, first check whether the equation is already set equal to zero — if not, move everything to one side first, since factoring and the zero product property only work on an equation equal to zero.
When solving a quadratic equation, try methods in this order of convenience: factor if the coefficients look nice (GCF, difference of squares, or a trinomial that factors with integers); complete the square if the equation is already in a form like $x^2+bx=c$ or the problem specifically asks for vertex form; use the quadratic formula whenever the expression won't factor over the integers, since it always works. The discriminant, $b^2-4ac$, tells you what kind of answer to expect before you even finish: positive means two real solutions, zero means one repeated real solution, negative means no real solutions.
For systems with one linear and one quadratic equation, solve for the same variable in both equations and set the expressions equal to each other — this turns the system into a single quadratic equation. Solve that quadratic (usually by factoring), then substitute each x-value back into the linear equation (it's simpler and less error-prone) to find the matching y-value. Always report solutions as ordered pairs, and remember a line can cross a parabola at 0, 1, or 2 points.
Finally, check every answer by substituting back into the original equation or expression — this catches sign errors and arithmetic slips before they cost points on the exam.
What you have to know
Zero Product Property
If $a\cdot b=0$, then $a=0$ or $b=0$. This only applies when one side of the equation is exactly 0.
Quadratic Formula
For $ax^2+bx+c=0$ with $a\neq0$, the solutions are $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.
Discriminant Rule
For $ax^2+bx+c=0$, the value $b^2-4ac$ determines the solutions: positive gives two real solutions, zero gives one real (repeated) solution, negative gives no real solutions.
Completing the Square
To turn $x^2+bx$ into a perfect square trinomial, add $(b/2)^2$: $x^2+bx+(b/2)^2=(x+b/2)^2$. Whatever is added to one side of an equation must also be added to the other.
Difference of Squares
$a^2-b^2=(a-b)(a+b)$, valid whenever a binomial is a subtraction of two perfect squares.
14 practice questions
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Multiply: $(2x-3)(x+5)$
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Answer. $2x^2+7x-15$
Distribute each term: $2x\cdot x=2x^2$, $2x\cdot5=10x$, $-3\cdot x=-3x$, $-3\cdot5=-15$. Combine like terms $10x-3x=7x$ to get $2x^2+7x-15$. -
Subtract: $(4x^2-3x+2)-(x^2+5x-6)$
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Answer. $3x^2-8x+8$
Distribute the negative sign to every term of the second polynomial: $4x^2-3x+2-x^2-5x+6$. Combine like terms: $4x^2-x^2=3x^2$, $-3x-5x=-8x$, $2+6=8$. -
Expand and simplify: $(x-4)^2$
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Answer. $x^2-8x+16$
Use the square-of-a-binomial pattern $(a-b)^2=a^2-2ab+b^2$ with $a=x$, $b=4$: $x^2-2(x)(4)+16=x^2-8x+16$. -
Factor completely: $3x^2+12x$
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Answer. $3x(x+4)$
The GCF of $3x^2$ and $12x$ is $3x$. Dividing each term by $3x$ leaves $x+4$, so the factored form is $3x(x+4)$. -
Factor completely: $x^2+2x-15$
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Answer. $(x+5)(x-3)$
Find two numbers that multiply to $-15$ and add to $2$: those are $5$ and $-3$. So the trinomial factors as $(x+5)(x-3)$. -
Factor completely: $16x^2-49$
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Answer. $(4x-7)(4x+7)$
Recognize this as a difference of squares, $a^2-b^2$, with $a=4x$ and $b=7$. Applying the pattern gives $(4x-7)(4x+7)$. -
Factor completely by grouping: $x^3-2x^2-9x+18$
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Answer. $(x-2)(x-3)(x+3)$
Group as $(x^3-2x^2)+(-9x+18)$, factor each group: $x^2(x-2)-9(x-2)$. Factor out $(x-2)$: $(x-2)(x^2-9)$, then factor the difference of squares to get $(x-2)(x-3)(x+3)$. -
Solve by completing the square: $x^2-4x-12=0$
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Answer. $x=6$ or $x=-2$
Move the constant: $x^2-4x=12$. Add $(-4/2)^2=4$ to both sides: $x^2-4x+4=16$, so $(x-2)^2=16$. Take square roots: $x-2=\pm4$, giving $x=6$ or $x=-2$. -
Write $y=x^2+6x+5$ in vertex form by completing the square.
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Answer. $y=(x+3)^2-4$
Take half of 6, square it: $(6/2)^2=9$. Add and subtract 9 inside the expression: $x^2+6x+9-9+5$. Group the perfect square: $(x+3)^2-4$. -
Solve using the quadratic formula: $3x^2+5x-2=0$
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Answer. $x=\dfrac{1}{3}$ or $x=-2$
Here $a=3,b=5,c=-2$. The discriminant is $5^2-4(3)(-2)=25+24=49$, and $\sqrt{49}=7$. So $x=\dfrac{-5\pm7}{6}$, giving $x=\dfrac{2}{6}=\dfrac13$ or $x=\dfrac{-12}{6}=-2$. -
What does the discriminant of $x^2-4x+13=0$ tell you about its solutions?
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Answer. Two non-real (complex) solutions, since the discriminant is negative
Here $a=1,b=-4,c=13$, so the discriminant is $(-4)^2-4(1)(13)=16-52=-36$. Since this is negative, the equation has no real solutions. -
Solve by factoring: $x^2-x-42=0$
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Answer. $x=7$ or $x=-6$
Find two numbers that multiply to $-42$ and add to $-1$: those are $-7$ and $6$, but checking signs, $7$ and $-6$ work since $7\times(-6)=-42$ and $7+(-6)=1$... correctly the factors are $(x-7)(x+6)=0$, giving $x=7$ or $x=-6$. -
What value of k makes $x^2-14x+k$ a perfect square trinomial?
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Answer. $k=49$
Take half the coefficient of x and square it: $(-14/2)^2=(-7)^2=49$. Adding 49 makes the expression $(x-7)^2$, a perfect square trinomial. -
Solve the system algebraically: $y=3x-2$ and $y=x^2+x-2$.
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Answer. $(0,-2)$ and $(2,4)$
Set the expressions equal: $x^2+x-2=3x-2$. Simplify to $x^2-2x=0$, then factor: $x(x-2)=0$, giving $x=0$ or $x=2$. Substitute into the linear equation $y=3x-2$: for $x=0$, $y=-2$; for $x=2$, $y=4$.
What people get wrong
- When subtracting one polynomial from another, students distribute the negative sign to only the first term of the second polynomial. Instead, rewrite the subtraction as adding the opposite of every term before combining like terms.
- When completing the square, students add $(b/2)^2$ to only one side of the equation, which breaks the equality. Instead, add the same value to both sides, or move the constant to the other side first and add there too.
- When factoring a trinomial with a leading coefficient other than 1 (like $2x^2+7x+3$), students apply the simple $(x+p)(x+q)$ pattern that only works when $a=1$. Instead, use the ac-method or check for a common factor first, and always verify by multiplying back.
- When applying the quadratic formula, students mishandle the negative sign on $b$, especially when squaring it, writing $b^2$ instead of correctly computing $(-b)^2$ or $b^2-4ac$ with a negative $b$. Instead, substitute all values in parentheses first, then simplify.
- In a linear-quadratic system, students find the x-values from the combined equation but forget to substitute back to find y, or they plug into the quadratic equation instead of the simpler linear one, increasing the chance of arithmetic error. Instead, always finish by substituting into the linear equation and reporting full ordered pairs.
- Students apply the zero product property to an equation that isn't set equal to zero, such as solving $(x-3)(x+2)=6$ by setting each factor equal to 6. Instead, expand and move everything to one side to get the equation equal to 0 before factoring.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Algebra I. Free, and no account needed to start.