ALGEBRA · 01

Factorization & the quadratic formula

A ball is thrown from a height. After $t$ seconds its height is given by an expression with a $t^2$ term, a $t$ term and a constant (the projectile page shows where it comes from). A sum like $x^2+5x+6$ is an expression; each piece is a term: $x^2$ , $5x$ , and the plain number $6$ , which is the constant term. An equation with an $x^2$ term, such as $x^2+5x+6=0$ , is called a quadratic equation. When does the ball reach the ground? We need the value of $t$ that makes the height zero, which means solving a quadratic equation. Usually we put an input into an expression and calculate the output, as on the function page. Here the question runs backwards: we know the output we want and must find the input that produces it.

Expanding $(x + 2)(x + 3)$ gives $x^2 + 5x + 6$ . Taking one term from each bracket gives $x\cdot x=x^2$ , $x\cdot 3=3x$ , $2\cdot x=2x$ and $2\cdot 3=6$ ; the middle terms add to $5x$ . Now cover the brackets and leave the expansion visible. To recover them, the two numbers must add to 5 and multiply to 6. We have something specific to look for. Rewriting a sum as this kind of product is factorization. If suitable integers do not come to mind, we need not keep guessing: later we will complete a square in the diagram, explain the corresponding algebra, and derive the quadratic formula from that construction.

01Where does every term come from?

The exponent 2 means two identical brackets, not squaring each term separately. Choose a formula below and play all four selections.

(a+b)^2

Selection 1

a\cdot a=a^2

Selection 2

a\cdot b=ab

Selection 3

b\cdot a=ab

Selection 4

b\cdot b=b^2

**Multiplication distributes**: choose one term from **each bracket**, multiply, and add all four possible products.

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(a+b)^2=a^2+2ab+b^2

x^2+6x+9=x^2+2\cdot3\cdot x+3^2=(x+3)^2

Read the identity backwards: x takes the role of a and 3 takes the role of b.

(a\pm b)^2=a^2\pm2ab+b^2

Use the matching sign throughout. The last term is always $+b^2$ , because both $b\cdot b$ and (−b)(−b) are $b^2$ . In contrast, $(a+b)(a-b)$ has a last term $-b^2$ .

02Use the expansion to find the factors.

For $ax^2+bx+c$ , suppose the factors are $(Ax+B)(Cx+D)$ . Capital $A$ , $B$ , $C$ , $D$ are unknown constants. Lowercase $a$ is the number in front of $x^2$ (the leading coefficient), $b$ the number in front of $x$ , and $c$ the constant. Multiply one term from each bracket, then collect the two terms containing $x$ . If two expressions are equal for every $x$ , their $x^2$ , $x$ and constant coefficients must match. Matching the squared, linear and constant terms gives the three conditions below.

(Ax+B)(Cx+D)=ACx^2+ADx+BCx+BD

(Ax+B)(Cx+D)=ACx^2+(AD+BC)x+BD

AC=a,\qquad AD+BC=b,\qquad BD=c

2x^2-2x-4

(x-2)(2x+2)

Choose factors for the leading coefficient and the constant. The crossing products must add to the middle coefficient. Cross-multiplying means multiplying $A$ with $D$ and $B$ with $C$ , then adding the two products. The display always shows the product you actually chose.

AC = 2BD = -4

Matched: -2 is the required coefficient -2. The factorization is valid.

2x^2-2x-4=(x-2)(2x+2)

1\cdot2=2,\quad(-2)\cdot2=-4,\quad1\cdot2+(-2)\cdot2=-2

If this expression equals zero, $(x-2)(2x+2)=0$ . Why must a factor be zero? If $x-2$ is not zero, we may divide by it, leaving $2x+2=0$ . Thus either $x-2=0$ , giving $x=2$ , or $2x+2=0$ , giving $x=-1$ . Substituting either answer makes the original product zero. Integer factors are convenient, but not every quadratic has them. We need another way to solve those equations: completing the square.

03Build the quadratic formula, one equality at a time.

A quadratic equation has the standard form $ax^2+bx+c=0$ , with $a\ne0$ . We want to isolate $x$ . Turning the left side into a square will let us take square roots. The area picture first shows why a corner must be added; the algebra then works for all signs of the coefficients. Taking a square root can give two signs: if $(x+3)^2=16$ , then $x+3$ is either $4$ or $-4$ . The notation $|a|$ means the size of $a$ without its sign. Click any line to inspect that step.

Area example: $x^2 + 6x$ . Split 6x into two strips of area 3x, and move one around the square. Here $x > 0$ .

x^2+6x=(x+3)^2-9

The missing corner is

3 \times 3 = 9

. Adding it completes a square of side

x+3

. For negative

b

, use the same algebra; do not interpret a negative number as a physical side length.

DERIVATION 1 / 9

ax^2+bx+c=0\quad(a\ne0)

The leading coefficient a is nonzero; otherwise this is not quadratic.

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What does the square root tell us?

\Delta=b^2-4ac=0

Zero $\Delta$ : the two signs give the same root, a repeated real root.

x=-3

Factorization & the quadratic formula

01Where does every term come from?

02Use the expansion to find the factors.

03Build the quadratic formula, one equality at a time.

What does the square root tell us?

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