Foundation 01
Multiplying Binomials & Factoring Quadratics
What it is & why you need it
Multiplying binomials means turning something like \((x+2)(x+3)\) into \(x^2+5x+6\).
Factoring is the reverse — going from \(x^2+5x+6\) back to \((x+2)(x+3)\).
In Module 2 this is the bridge between the factored form of a quadratic and its standard form,
and factoring is how you find where a parabola crosses the x-axis.
Worked example — multiply then factor
Multiply \((x+2)(x+3)\) using FOIL (First, Outer, Inner, Last):
- First: \(x \cdot x = x^2\)
- Outer: \(x \cdot 3 = 3x\)
- Inner: \(2 \cdot x = 2x\)
- Last: \(2 \cdot 3 = 6\)
- Add them up and combine the like terms: \[x^2 + 3x + 2x + 6 = x^2 + 5x + 6\]
Now factor \(x^2+5x+6\) by finding two numbers that multiply to 6 and add to 5. Those numbers are \(2\) and \(3\), so:
\[x^2 + 5x + 6 = (x+2)(x+3)\]
Try it
1. Multiply \((x+4)(x+1)\).
Show answer
First \(x\cdot x = x^2\); Outer \(x\cdot 1 = x\); Inner \(4\cdot x = 4x\); Last \(4\cdot 1 = 4\).
Combine: \[x^2 + x + 4x + 4 = \mathbf{x^2 + 5x + 4}\]
2. Factor \(x^2 + 7x + 12\).
Show answer
Find two numbers that multiply to \(12\) and add to \(7\): those are \(3\) and \(4\).
\[x^2 + 7x + 12 = \mathbf{(x+3)(x+4)}\]
Check by multiplying back: \((x+3)(x+4) = x^2+7x+12.\) ✓
Foundation 02
Square Roots
What it is & why you need it
A square root asks, "what number times itself gives this?" Since \(5\times 5 = 25\),
we say \(\sqrt{25} = 5\). In Module 2 you'll undo squaring to solve equations like \(x^2 = 9\), and
square roots are the engine inside the quadratic formula. Knowing your perfect squares makes this fast.
Worked example — solve a squared equation
Solve \(x^2 = 49\):
- Squaring is undone by taking a square root of both sides.
- Ask: what number squared is \(49\)? Since \(7^2 = 49\), one answer is \(7\).
- Don't forget the negative. \((-7)^2 = 49\) too, so we write \(\pm\):
- \[x = \pm\sqrt{49} = \pm 7 \quad\Rightarrow\quad x = 7 \text{ or } x = -7\]
Helpful to memorize: \(\sqrt{1}=1,\ \sqrt{4}=2,\ \sqrt{9}=3,\ \sqrt{16}=4,\ \sqrt{25}=5,\ \sqrt{36}=6,\ \sqrt{49}=7,\ \sqrt{64}=8,\ \sqrt{81}=9,\ \sqrt{100}=10.\)
Try it
1. What is \(\sqrt{36}\)?
Show answer
We need the number that times itself gives \(36\). Since \(6\times 6 = 36\), \[\sqrt{36} = \mathbf{6}.\]
2. Solve \(x^2 = 16\).
Show answer
Take the square root of both sides and keep both signs:
\[x = \pm\sqrt{16} = \pm 4 \quad\Rightarrow\quad \mathbf{x = 4 \text{ or } x = -4}.\]
Foundation 03
Solving Quadratics by Factoring (Algebra I)
What it is & why you need it
This combines Foundations 01 and 02. To solve a quadratic like \(x^2 + 5x + 6 = 0\), you factor it and
then use the Zero Product Property: if two things multiply to \(0\), at least one of them
must be \(0\). In Module 2 the solutions you find this way are exactly the roots —
the spots where the parabola touches the x-axis.
Worked example
Solve \(x^2 + 5x + 6 = 0\):
- Factor the left side (two numbers that multiply to \(6\), add to \(5\) → \(2\) and \(3\)): \[(x+2)(x+3) = 0\]
- Zero Product Property: set each factor equal to \(0\). \[x+2 = 0 \quad\text{or}\quad x+3 = 0\]
- Solve each little equation: \[x = -2 \quad\text{or}\quad x = -3\]
- So the roots are \(x = -2\) and \(x = -3\). These are where the parabola crosses the x-axis.
Try it
1. Solve \(x^2 + 7x + 12 = 0\).
Show answer
Factor (multiply to \(12\), add to \(7\) → \(3\) and \(4\)): \((x+3)(x+4)=0.\)
Set each factor to zero: \(x+3=0\) or \(x+4=0\), giving \[\mathbf{x = -3 \text{ or } x = -4}.\]
2. Solve \(x^2 - 9 = 0\). (Hint: this is a difference of squares, \(x^2-9=(x-3)(x+3)\).)
Show answer
Factor: \((x-3)(x+3)=0.\) Set each factor to zero: \(x-3=0\) or \(x+3=0\), so
\[\mathbf{x = 3 \text{ or } x = -3}.\]
Foundation 04
Exponents
What it is & why you need it
An exponent is a shortcut for repeated multiplication: \(x^2\) means \(x\cdot x\), and
\(3^2\) means \(3\cdot 3 = 9\). The little \(2\) is exactly what makes a function "quadratic." In Module 2
you'll square numbers all the time — especially when plugging values into \(f(x)=x^2\) — so getting
comfortable with squaring (including negatives) pays off everywhere.
Worked example — squaring carefully
Evaluate \((-4)^2\):
- The exponent \(2\) means multiply the base by itself: \((-4)^2 = (-4)\cdot(-4)\).
- A negative times a negative is a positive.
- So \[(-4)\cdot(-4) = +16.\]
- Careful: \(-4^2\) without parentheses means \(-(4^2) = -16\). The parentheses matter!
Try it
1. Evaluate \(5^2\).
Show answer
\(5^2 = 5\cdot 5 = \mathbf{25}.\)
2. Evaluate \((-3)^2\) and then \(2^3\).
Show answer
\((-3)^2 = (-3)\cdot(-3) = \mathbf{9}\) (negative times negative is positive).
\(2^3 = 2\cdot 2\cdot 2 = \mathbf{8}.\)
Foundation 05
Evaluating Functions
What it is & why you need it
"Evaluating a function" just means plugging a number in for \(x\) and simplifying.
The notation \(f(3)\) reads "the value of \(f\) when \(x = 3\)." In Module 2 you'll evaluate quadratic
functions to build tables, find the y-intercept, and check whether a point is on the parabola. It's the
workhorse skill behind every graph.
Worked example
If \(f(x) = x^2 + 2x - 1\), find \(f(3)\):
- Wherever you see \(x\), substitute \(3\) (use parentheses to stay safe): \[f(3) = (3)^2 + 2(3) - 1\]
- Handle the exponent first: \((3)^2 = 9\).
- Then the multiplication: \(2(3) = 6\).
- Add and subtract left to right: \[9 + 6 - 1 = 14\]
- So \(f(3) = 14\). The point \((3,\,14)\) is on this parabola.
Try it
1. If \(f(x) = x^2 - 4\), find \(f(2)\).
Show answer
Substitute: \(f(2) = (2)^2 - 4 = 4 - 4 = \mathbf{0}.\) (So \(x=2\) is a root!)
2. If \(g(x) = x^2 + 3x\), find \(g(-2)\).
Show answer
Substitute with parentheses: \(g(-2) = (-2)^2 + 3(-2) = 4 + (-6) = \mathbf{-2}.\)
Foundation 06
The Coordinate Plane
What it is & why you need it
The coordinate plane is the grid where graphs live. Every point is an ordered pair
\((x,\,y)\): the \(x\) tells you how far right (or left, if negative), the \(y\) tells you how far
up (or down). In Module 2 the whole parabola is just a collection of these points — the vertex,
the roots, and the y-intercept are all specific \((x,\,y)\) locations you'll plot and read.
Worked example — plot a point
Plot the point \((-2,\,3)\):
- Always start at the origin \((0,\,0)\), the center where the axes cross.
- The first number is \(x = -2\): move 2 units left (negative is left).
- The second number is \(y = 3\): from there move 3 units up (positive is up).
- That landing spot is the point \((-2,\,3)\). It sits in the upper-left region (Quadrant II).
Remember the order: x first, then y — like reading "across, then up."
Try it
1. Which way do you move for the point \((4,\,-1)\)?
Show answer
From the origin: \(x=4\) means 4 units right, then \(y=-1\) means 1 unit down.
The point lands in the lower-right region (Quadrant IV).
2. A parabola has its lowest point at \((0,\,0)\) and also passes through \((1,\,1)\) and \((-1,\,1)\). Where is the y-intercept?
Show answer
The y-intercept is where the graph crosses the y-axis — that's where \(x = 0\). Here that point is
\(\mathbf{(0,\,0)}\), which is also the vertex. (This is the parent parabola \(y = x^2\).)