How to Factor Quadratics Using Algebra Tiles

Factoring quadratics using algebra tiles means arranging the tiles of a quadratic into one complete rectangle. The length and width of that rectangle are the factors. For \(x^2 + 5x + 6\), the 1 x²-tile, 5 x-tiles and 6 unit tiles fit into a rectangle with edges \(x + 2\) and \(x + 3\), so \(x^2 + 5x + 6 = (x + 2)(x + 3)\).

Arrange the tiles yourself: Try it with our free Algebra Tiles tool. Its factor mode checks whether your tiles make a full rectangle and labels the edges.

The tiles you need

• x²-tile: a large square worth \(x^2\).
• x-tile: a rectangle worth \(x\). Its long side matches the side of the x²-tile.
• Unit tile: a small square worth \(1\). Its side matches the short side of the x-tile.

Positive tiles are plain (white, yellow, green or blue) and shaded (or red) tiles are negative. Factoring is multiplying backward: in the area model for multiplying binomials you know the edges and find the tiles inside; here you know the tiles and find the edges.

Step by step: factor x² + 5x + 6 with algebra tiles

We will factor \(x^2 + 5x + 6\). That is 1 x²-tile, 5 x-tiles and 6 unit tiles, all positive.

1. Gather the tiles. Lay out 1 x²-tile, 5 x-tiles and 6 unit tiles.

   

2. Start in the corner. Put the x²-tile in the top-left corner. The unit tiles always go in the opposite (bottom-right) corner.
3. Arrange the unit tiles in a block. Six units can make a 1 by 6 block or a 2 by 3 block. Pick the block whose two side lengths add to the number of x-tiles. \(1 + 6 = 7\) is too many, but \(2 + 3 = 5\) matches.
4. Fill the gaps with x-tiles. With the units in 3 rows of 2, place 2 x-tiles standing up to the right of the x²-tile and 3 x-tiles lying down under it. All 5 x-tiles are used and the shape is a complete rectangle.

   

5. Read the edges. Across the top you see one x and two units, \(x + 2\). Down the left side you see one x and three units, \(x + 3\).

   

6. Write and check. \(x^2 + 5x + 6 = (x + 2)(x + 3)\). Check by multiplying: \(x^2 + 3x + 2x + 6 = x^2 + 5x + 6\).
7. Try it. Open the free Algebra Tiles tool, choose factor, enter \(x^2 + 5x + 6\) and try both unit blocks. Only the 2 by 3 block closes the rectangle.

More worked examples

Example 1: x² + 7x + 12

Twelve units can be 1 by 12, 2 by 6 or 3 by 4. The side lengths must add to 7, and only \(3 + 4 = 7\) works. Place 3 x-tiles on one side of the x²-tile and 4 on the other: \(x^2 + 7x + 12 = (x + 3)(x + 4)\).

Example 2: a perfect square, x² + 6x + 9

Nine units fit in a 3 by 3 block, and \(3 + 3 = 6\) matches the x-tiles. Put 3 x-tiles on each side. Both edges are \(x + 3\), so the rectangle is a square: \(x^2 + 6x + 9 = (x + 3)^2\). When the tiles make a square, you have a perfect square trinomial.

Example 3: a common factor, 2x² + 6x

There are no unit tiles. Stack the 2 x²-tiles one above the other and put 3 x-tiles beside each. The left edge is \(x + x = 2x\) and the top edge is \(x + 3\): \(2x^2 + 6x = 2x(x + 3)\).

Example 4: negatives and zero pairs, x² + x − 6

The tiles are 1 x²-tile, 1 positive x-tile and 6 shaded unit tiles. One x-tile cannot fill both gaps, so add zero pairs of x-tiles, which do not change the value.

1. Put the 6 shaded units in a 2 by 3 block. The two gaps need 3 x-tiles on one side and 2 on the other.
2. Add 2 zero pairs of x-tiles. You now have 3 positive and 2 shaded x-tiles, still worth \(3x – 2x = x\).
3. Put the 3 positive x-tiles in the gap that touches the positive edge and the 2 shaded x-tiles in the gap that touches the negative edge. The edges read \(x + 3\) and \(x – 2\).

So \(x^2 + x – 6 = (x + 3)(x – 2)\). Check: \(x^2 – 2x + 3x – 6 = x^2 + x – 6\).

Example 5: leading coefficient 2, 2x² + 5x + 3

Put the 2 x²-tiles side by side, so the top edge starts with \(2x\). Three units can be 1 row of 3 or 3 rows of 1. With 1 row of 3 units in the bottom-right corner, you need 3 x-tiles standing on the right and 2 x-tiles lying under the x²-tiles, which is \(3 + 2 = 5\). It works: \(2x^2 + 5x + 3 = (2x + 3)(x + 1)\). The other block would need \(1 + 6 = 7\) x-tiles.

Example 6: when no rectangle exists, x² + 5x + 5

Five units only fit as 1 by 5, which needs \(1 + 5 = 6\) x-tiles, and adding zero pairs cannot change that total. No complete rectangle is possible, so \(x^2 + 5x + 5\) is prime over the integers: it does not factor into binomials with whole-number terms.

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Common mistakes when factoring with tiles

• Accepting a shape with gaps. Every tile must be used and the outline must be a full rectangle. An L-shape or a missing corner means the arrangement is wrong.
• Picking the unit block by product only. Both 1 by 6 and 2 by 3 multiply to 6. The block must also make the x-tile count work.
• Reading the wrong edges. The factors are the edge lengths, not the tiles inside. Read one full side across the top and one full side down the left.
• Mixing up signs with negatives. In \(x^2 + x – 6\), the shaded x-tiles go next to the negative edge. Check by multiplying: \((x – 3)(x + 2)\) gives \(x^2 – x – 6\), not \(x^2 + x – 6\).
• Forgetting the common factor. With \(2x^2 + 6x\), the left edge is \(2x\), not just 2.

Practice exercises with answers

Arrange each set of tiles into a rectangle and read the edges.

1. \(x^2 + 4x + 3\)
2. \(x^2 + 6x + 8\)
3. \(x^2 + 4x + 4\)
4. \(3x^2 + 6x\)
5. \(x^2 + 2x – 3\)
6. \(2x^2 + 7x + 3\)

Answers:

1. \((x + 1)(x + 3)\)
2. \((x + 2)(x + 4)\), using the 2 by 4 unit block
3. \((x + 2)^2\), a square
4. \(3x(x + 2)\)
5. \((x + 3)(x – 1)\), after adding 1 zero pair of x-tiles
6. \((2x + 1)(x + 3)\)

Factoring quadratics with algebra tiles practice quiz: 10 questions

Test yourself with 10 multiple-choice questions. Choose an answer and press Submit to see if you are right and why. Your score appears at the end.

Frequently asked questions

How do you factor with algebra tiles?

Model the quadratic with x²-tiles, x-tiles and unit tiles. Put the x²-tiles in a corner, the unit tiles in a block in the opposite corner, and the x-tiles in the gaps until you have one complete rectangle. The expressions along the top and side edges are the two factors.

How do you factor trinomials using algebra tiles?

For \(x^2 + bx + c\), try each block of c unit tiles whose side lengths multiply to c. The right block is the one whose side lengths add to b, because those are the x-tiles you need to fill the gaps. For \(x^2 + 5x + 6\), the 2 by 3 block works and the answer is \((x + 2)(x + 3)\).

How do you factor perfect square trinomials with algebra tiles?

Arrange the tiles as usual. If the finished rectangle is a square, the trinomial is a perfect square. For \(x^2 + 6x + 9\), the 9 units form a 3 by 3 block, 3 x-tiles go on each side, and both edges are \(x + 3\), so it equals \((x + 3)^2\).

Is factoring quadratics Algebra 1 or Algebra 2?

Both. In most US courses, factoring quadratics like \(x^2 + 5x + 6\) is first taught in Algebra 1, often with algebra tiles. Algebra 2 returns to it with harder cases, such as larger leading coefficients and higher-degree polynomials. Tiles are most useful for the Algebra 1 cases with small numbers.

What if the rectangle doesn't come out even?

First try the other unit blocks, and if the trinomial has negative terms, add zero pairs of x-tiles. If no arrangement closes the rectangle, the quadratic does not factor over the integers. \(x^2 + 5x + 5\) is an example: it is prime, and you would use the quadratic formula to find its roots.

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