Factoring quadratics with an area model

• Relate the terms of a quadratic trinomial to partial areas in a rectangle.
• Choose two integers whose product and sum match the constant and linear coefficients.
• Verify a factorization by distribution and by checking the resulting zeros.

Build the idea

Worked example

1. Step 1

   The positive factor pairs of six are one and six, or two and three. Only two and three also add to the required middle coefficient five.

   $$2 \cdot 3 = 6 \quad\text{and}\quad 2 + 3 = 5$$

2. Step 2

   Rewrite five x as two x plus three x. The polynomial now names the four regions of the area model without changing their total area.

   $$x^2 + 2x + 3x + 6$$

3. Step 3

   Arrange the x-squared square, the two strips, and the constant corner into a rectangle. Its side lengths are x plus two and x plus three.

   $$x^2 + 5x + 6 = (x + 2)(x + 3)$$

4. Step 4

   Distribute each term from the first factor across the second. Combining the two middle products reproduces five x, so the proposed factorization is equivalent.

   $$(x + 2)(x + 3) = x^2 + 3x + 2x + 6$$

• Choosing any factor pair of six without also checking that the pair adds to five.

  The constant coefficient controls the product, while the linear coefficient controls the sum. A valid pair must satisfy both conditions; one condition alone does not reconstruct the trinomial.

• Writing factors with negative signs because the zeros of the quadratic are negative.

  The factors are x plus two and x plus three. Their zeros are negative two and negative three because each factor becomes zero after subtracting its positive constant from both sides.

Use a sum-and-product check to factor x squared plus seven x plus twelve, then verify by distribution.

This reviewed reading order covers the lesson explanation, equations, example, and checks without claiming that a video exists.

1. Part 1

   Factoring reverses expansion

   Start with x squared plus five x plus six. Instead of viewing it as three disconnected terms, imagine a total area assembled from simpler regions. Factoring asks for the outer side lengths of one rectangle whose parts add to this polynomial. Expanding those side lengths later will test whether our reconstruction is correct.

   • x squared plus five x plus six.
2. Part 2

   Choose the two strip widths

   The constant corner must have area six, so its side lengths multiply to six. The two x strips must combine to five x, so those same side lengths add to five. Two and three satisfy both requirements. One and six have the right product but the wrong sum, so they cannot complete this rectangle.

   • Two times three equals six, and two plus three equals five.
3. Part 3

   Read the factors

   Split five x into two x and three x. Place those strips along adjacent sides of the x-by-x square, then place the two-by-three constant corner in the remaining space. The full horizontal side is x plus two, and the full vertical side is x plus three. Their product is the factored form.

   • x squared plus two x plus three x plus six.
   • x squared plus five x plus six equals the quantity x plus two times the quantity x plus three.
   • The quantity x plus two times the quantity x plus three.
4. Part 4

   Multiply back to check

   Distribute x and two across the second factor. The four products are x squared, three x, two x, and six. Combining the two middle terms returns five x, exactly matching the starting trinomial. This reverse trip confirms both the numbers and their signs rather than trusting a pattern match.

   • The quantity x plus two times the quantity x plus three equals x squared plus three x plus two x plus six.
   • x squared plus five x plus six.

Scope and limitations

• This lesson treats monic quadratics that factor over integers and does not present the quadratic formula or irrational and complex roots.
• An area model is most direct for positive dimensions; signed coefficients require algebraic tiles or symbolic grouping to extend the picture carefully.