6.01d, 6.01e - Quadratic expressions

Multiplying two binomials

A rectangle has side lengths $x+3$ and $x+2$. Is its area $x^2+6$? Splitting each side shows two extra regions that this tempting answer misses.

[DIAGRAM: asset_name: Four products in a rectangle; asset_slug: two-bracket-area; caption: Each term in the first bracket multiplies each term in the second.; description: A rectangle of width x plus three and height x plus two, split into four regions. The widths are x and three; the heights are x and two. The four areas are labelled x squared, three x, two x and six. The picture uses positive x.]

The four areas add to $x^2+3x+2x+6$. The two middle regions explain why multiplying only the first terms and the last terms is incomplete. A binomial is an expression with two terms; multiplying two binomials gives four products before any like terms are collected.

Use distribution twice: first $a(c+d)+b(c+d)$, then distribute within each part. A grid is another way to keep track of the same four products.

Change only the sign of the two in the second bracket. Two of the four products now change sign.

The same method works with other letters or coefficients. For example, $(a+2b)(a-b)=a^2-ab+2ab-2b^2=a^2+ab-2b^2$. The negative sign belongs to $-b$ in both of its products.

Use the same structure with a fresh positive and negative constant.

Finding quadratic factors

Expanding the general product $(x+p)(x+q)$ gives $x^2+(p+q)x+pq$. To reverse that process, find two numbers whose sum gives the middle coefficient and whose product gives the constant term.

When seeking integer factors, list factor pairs of $|c|$ and use the signs. A positive product requires two signs the same; their sum decides which sign. A negative product requires opposite signs; their sum decides which number has the greater magnitude. If no integer pair works, say so rather than forcing a false factorisation. For example, $x^2+x+1$ has no such integer pair: the only integer pairs with product one are $(1,1)$ and $(-1,-1)$, whose sums are two and negative two, not one.

Change only the sign of the constant term. The product must now be negative, so the two numbers have opposite signs.

A repeated pair is allowed too: $x^2-6x+9=(x-3)^2$. If the constant is zero, use the common-factor method: $x^2+4x=x(x+4)$.

The next quadratic does have an integer pair; use both conditions to find it.

Differences of squares

Multiply $(a-b)(a+b)$ and the middle products cancel: one is $+ab$, the other is $-ab$. This makes a two-term expression factorise into two binomials.

Recognise both squares before writing the factors. For example, $x^2-25=x^2-5^2$, so its factors are $(x-5)(x+5)$. You need subtraction: $x^2+25$ does not follow this difference-of-squares pattern.

Change only the constant from nine to five. Five is not the square of an integer, but it is the square of its exact square root. You do not need a decimal approximation.

This is the connection: “not a square number” does not mean “cannot be written as a square”. Every positive real number $c$ is $(\sqrt c)^2$, so $x^2-c=(x-\sqrt c)(x+\sqrt c)$. At $c=0$, this becomes $x\times x$. For $c<0$, $\sqrt c$ is not real, so that real-root construction does not apply.

Keep the square root exact in this new factorisation.

Quick Recap

• Multiply every term in one binomial by every term in the other, then collect like terms.
• For monic quadratic factors, the two constants must have both the correct sum and the correct product.
• Not every monic quadratic has integer factors.
• In a difference of squares, opposite-sign brackets cancel the middle terms.
• Exact square roots let you factorise $x^2-c$ for positive $c$ without rounding.

The rectangle's area is $x^2+5x+6$, because the two middle regions contribute $3x+2x$. Multiplying only the corners misses part of the area.