# Think Inside the Box (Part 2): Using the Box with Polynomials

The “box” or tabular method of multiplication is a great way for elementary, middle, and high school students to make mathematical connections. In this second of two articles, I discuss how the box method can be used to multiply and divide polynomials.

(In the first article, I discussed how the box method can be used to multiply and divide numbers.)

This article expands on ideas presented at a Math for America workshop that I co-facilitated in 2017 in New York City.

## Multiplying polynomials

8. (x + 4)(x + 3)

We multiply polynomials using the box method in the same way that we multiply whole numbers. We create a separate row or column for each term in a factor, multiply the dimensions of each cell, and add like terms:

x2 + 4x + 3x + 12 = x2 + 7x + 12

The box method is actually a more abstract representation of using algebra tiles to construct a rectangle:

9. (3x2 – 2x + 4)(x + 3)

In this case, we need a 3 x 2 rectangle:

= 3x3 + 9x2 – 2x2 – 6x + 4x + 12 = 3x3 + 7x2 – 2x + 12

## Dividing polynomials

10. (x2 – 10x + 25) ÷ (x – 4)

We can divide polynomials using the box method in the same way that we divide whole numbers.

In this example, we need a 2 x 2 rectangle. We know from #2 that since the dividend has three terms, then the resulting rectangle must have three diagonals. We know that we must have two rows and two columns in the rectangle since the divisor x – 4 has two terms, so we can set up the following rectangle:

We know that the sum of the diagonals is the dividend x2 – 10+ 25:

We know that x2 goes in the top left box (m), since there is only one box that represents the x2 term of the dividend.

Now we work in reverse: what times x equals x2? (or, what is x2 ÷ x?) We write the answer, x:

Next, we multiply –4(x) to get –4x, which is the value of n:

Repeating the process, we can subtract –10x – (–4x) to find the value of p, so p = –6x. We can divide –6x ÷ x = –6 = b:

Then we multiply –6(–4) = 24 = q.

Since the sum of the yellow diagonal is 25, then there must be a remainder of 25 – 24 = +1.

Thus, (x2 – 10x + 25) ÷ (x – 4) = 1/(x – 4).

11. (x4 +4x3 + 3x2 + 4x – 5) ÷ (x2 + 1)

The divisor here has two terms, but adding a placeholder of 0x in the divisor will keep the method consistent with dividing whole numbers (just as we would say that 1(102) + 0(101) + 1(100) = 101, then x2 + 1 = x2 + 0x + 1) and will help line up the like terms properly in the rectangle.

Since the dividend has 5 terms (one for each diagonal) and the divisor has three terms, then we create a table with three rows (one for each row of the diagonal):

Starting at the top left of the rectangle, we see that x4 goes in the top left since it is the first term in the dividend. Dividing x4/x2 gives us x2 as the first term of the dividend. Multiplying the remaining terms in the divisor by x2 gives us 0x and x2:

Subtracting 4x3 – 0x3 gives us the topmost box in the second column, 4x3. Continuing our divide/multiply/subtract/bring down algorithm gives us the following:

Note that the last diagonal (in tan) shows a remainder of –7, obtained by subtracting –5 – (+2). The final answer is x2 + 4x + 2 – 7/(x2 + 1).

## Factoring trinomials

12. Factor x2 + 11x + 30.

Factoring differs from dividing because we know that there is no remainder and we are not given the divisor. In this case, we need the two binomial factors m + n and p + q that multiply together to equal the given trinomial x2 + 11x + 30. Since the factors are binomials, we can construct a 2 x 2 rectangle and write the diagonal sums on the bottom:

We find the first terms of the factors. Since mp = x2, then m = x and p = x:

We now need to find the two quantities n and q whose sum is +11x and product is +30. Here, the quantities in question are +5x and +6x, so the missing term in each factor is +5x/x = +5 and +6x/x = +6, giving us:

Thus, our answer is (x + 6)(x + 5).

13. Factor 6x2 + 10x – 9x – 15.

The box method provides a nice visual justification for factoring by grouping, which is the method typically used to factor this expression. We can represent the first two terms with a 2 x 1 rectangle and the second two terms with a 2 x 1 rectangle. Both rectangles must have the same width (i.e. a common factor):

To find p and q, we factor the greatest common factor m and n from each binomial:

We can then find p and q by dividing, i.e. 6x2/2x = 3x, 10x/2x = 5, –9x/–3 = 3x, and –15/–3 = 5:

The width of each rectangle (i.e. the factor 3x + 5) must be common to both columns in order to combine the two columns to make a rectangle. Thus, the answer is (2x – 3)(3x + 5).

14. Factor 8x2 – 37x – 15.

To factor quadratic trinomials whose leading coefficient is not equal to 1, we can use the “ac” method: for a trinomial ax2 + bx + c, where a ≠ 0, find the product ac. Then find the two factors of ac whose sum is the middle coefficient b. These numbers a and c are the coefficients of the two terms that the middle term bx can be split into.

In this example, a = 8, b = –37, c = –15. Since ac = 8(–15) = –120, then we need the factors of –120 that add up to –37, i.e. –40 and +3. Thus, 8x2 – 37x – 15 = 8x2 – 40x + 3x – 15. Using the box method, we find the greatest common factor for each pair of terms and write it on the top of each column, then divide:

Our answer is (8x + 3)(x – 5).

We can prove the ac method using a rectangle. Given ax2 + bx + c, where a ≠ 0. Let m and n be numbers such that m + n = b and mn = ac. Then c = mn/a, so

We can represent this using a 2 x 2 rectangle:

Completing the square

15. Solve x2 + 6x + 2 = 0 by completing the square.

The box method also provides an excellent visual method for completing the square, which students often have difficulty doing with pure algebra.

To solve by completing the square, we subtract the constant term from both sides to move it over to the right side, giving us x2 + 6x = –2. We can then construct a 2 x 2 square (note that this must be a square, not just a rectangle) to represent the trinomial on the left side of the equation. We need a binomial x + n such that (x + n)2 = x2 + 6x = –2:

We can see by taking the square root of the first term x2 that the first term in the binomial is x. We can also see that n = 6/2 = 3 since nx + nx = 2nx = 6x:

To complete the square on the left side (note how the box method illustrates where the term comes from!), the lower right box must be 3(3) = 9. We have to add +9 to the right side of the equation, giving us:

Thus, (x + 3)2 =7. Taking the square root of both sides gives us x + 3 = ±√7, or x = –3 ±√7.

16. Find the center and radius of the circle whose equation is x2 + 10x + y2 – 8y – 8 = 0.

Moving the constant –8 over to the right side gives the equation x2 + 10x + y2 – 8y = 8. We can draw two 2 x 2 squares and complete the square for x2 + 10x and y2 – 8y – 8 separately:

Dividing 5x/x and –4y/y gives us +5 and –4, respectively, so the binomials that are squared must be x + 5 and y – 4:

To complete the squares, we add +25 to the first square, +16 to the second square, and +41 to the right side of the equation:

Thus, the equation is (x + 5)2 + (y – 4)2 = 49. From this equation, we can see that the center is (–5, 4) and the radius is 7.

## Conclusion

The box method has several advantages.

First, the box method helps students line up numbers. Many students have difficulty organizing their work. For example, most of my students get lost in their work when they do long division. In my experience, most students have cleaner work using the box method than using other methods.

Second, it gives students a motivation to move beyond manipulatives. While the box method can be represented concretely using algebra tiles, it shows students that they will have to think more abstractly in order to solve harder problems. (Representing x2 + 23x + 120 with algebra tiles would be tedious!)

Most importantly, it helps students make connections across a number of topics spanning several years. Instead of learning different techniques for multiplying numbers, dividing numbers, multiplying polynomials, factoring polynomials, and completing the square, students can use one method to solve problems throughout elementary, middle, and high school. As a result, students can see that math is not a collection of seemingly unrelated tricks, but an interconnected logical system of ideas.

If you have any corrections, suggestions, or other comments, let me know using the form below. Thanks!