# Synthetic Division With Coefficient Not 1 (Or A Quadratic Divisor)

Synthetic division can make life easier when you are dividing polynomials. However, it is only useful in certain specific situations, such as when dividing by a linear monic polynomial.

So, can you use synthetic division with a coefficient that is not 1?  You need a monic linear divisor to use synthetic division. That means the coefficient of x must be 1. However, you can divide by a linear divisor whose leading coefficient is not 1 if you do it in multiple steps. You can also divide by a quadratic divisor by using synthetic division repeatedly.

Of course, synthetic division is a useful shortcut, but its use is limited by the restriction on the leading coefficient (the coefficient of x must be 1).

In this article, we’ll talk about how to get around this restriction to use synthetic division to divide by a linear polynomial that is not monic.  We’ll also look at how to divide by a quadratic divisor by using synthetic division repeatedly.

Let’s get started.

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## Synthetic Division With A Coefficient That Is Not 1

Remember that synthetic division is used to divide any polynomial by a linear monic polynomial.  A linear monic polynomial has the form x + b, where b is a real number (b can be positive, zero, or negative).

Synthetic division is a shortcut for polynomial long division.  It helps us to avoid writing variables in the intermediate steps.

However, this raises the question of whether we can use synthetic division with a coefficient that is not 1.  The answer is: we can make it work, but we need to break the problem into three steps:

• 1.) Factor the divisor into a product of a constant and a linear monic polynomial.  That is, for the divisor ax + b, we would factor it as a(x + b/a).
• 2.) Use synthetic division to divide the dividend polynomial by the linear monic x + b/a.  Since the divisor x + b/a has a coefficient of 1, synthetic division will work as usual.
• 3.) Take the result from step 2 and divide the quotient (the resulting polynomial) by a.

These steps will work because dividing by x + b/a and then dividing the result by a is equivalent to dividing by ax + b.

If you are unsure about why this works, think about it this way.  If you want to divide a number by 6, you can instead divide it by 2, and then divide the result by 3.

The reason this works is that 6 = 2*3 (2 and 3 are the factors of 6).  Let’s see an example:

### Example: Dividing By 6 In Two Different Ways

Let’s say we want to divide 120 by 6.  First, we’ll divide by 2 and then 3:

• 120 / 2 = 60
• 60 / 3 = 20

So our answer is 20.  This is the same answer we get if we just divide by 6: 120 / 6 = 20.

So, dividing 120 by 2 and then dividing the result by 3 is equivalent to dividing by 6.

Let’s look at an example of how this works for a divisor with a coefficient that is not 1.

### Example: Synthetic Division When The Coefficient Is Not 1

Let’s say we want to divide x2 + 13x + 40 by 2x + 10.

Following the 3-step process above:

• 1.) First, we factor 2x + 10 as 2(x + 5).  [Note that this leaves us with a linear monic polynomial in parentheses – that is, the coefficient of x is 1].
• 2.) Next, we divide x2 + 13x + 40 by x + 5, using synthetic division.  This gives us a result of x + 8.
• 3.) Last, we divide the result x + 8 by 2.  This gives us a final answer of x/2 + 4.

We can verify this result if we FOIL (2x + 10)(x/2 + 4) to get x2 + 13x + 40.

Below, we see the details of step 2 (synthetic division) in this process:

We can also verify that this works the same way as polynomial long division to give us the correct answer.

## Can You Do Synthetic Division With A Fraction?

You can do synthetic division with a fraction if you introduce additional steps, as we outlined above.  We still need to factor the divisor so that one of the factors is a linear monic polynomial.

Let’s take a look at an example of how this would work:

### Example: Synthetic Division When The Coefficient Is A Fraction

Let’s say we want to divide 2x2 + 9x – 18 by (1/3)x + 2.

Following the 3-step process above:

• 1.) First, we factor (1/3)x + 2 as (1/3)(x + 6).  [Note that this leaves us with a linear monic polynomial in parentheses – that is, the coefficient of x is 1].
• 2.) Next, we divide 2x2 + 9x – 18 by x + 6, using synthetic division.  This gives us a result of 2x – 3.
• 3.) Last, we divide the result 2x – 3 by 1/3 (division by 1/3 is equivalent to multiplication by 3).  This gives us a final answer of 6x – 9.

We can verify this result if we FOIL ((1/3)x + 2)(6x – 9) to get 2x2 + 9x – 18.

Below, we see the details of step 2 (synthetic division) in this process:

We can also verify that this works the same way as polynomial long division to give us the correct answer.

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## Synthetic Division With A Quadratic Divisor

As we saw earlier, we can use synthetic division with divisors that do not have a leading coefficient on 1.  It just takes some extra steps.

The same idea applies to synthetic division with a quadratic divisor.  It is possible, but there are some extra steps involved.

The key is that we must use synthetic division repeatedly.  Here are the steps:

• 1.) Factor the quadratic divisor into a product of a constant and a quadratic monic polynomial.  That is, for the divisor ax2 + bx + c, we would factor it as a(x2 + (b/a)x + (c/a)).
• 2.) Factor the quadratic monic polynomial into a product of two linear monic polynomials.  The result will look something like this: a(x – r)(x – s), where r and s are the roots (solutions) of the quadratic.
• 3.) Use synthetic division to divide the dividend polynomial by the linear monic x – r.  Since the divisor x – r has a coefficient of 1, synthetic division will work as usual.
• 4.) Take the result from step 3 and divide it by x – s, using synthetic division again.
• 5.) Take the result from step 4 and divide by a.

This is equivalent to dividing by ax2 + bx + c since a(x – r)(x – s) = a(x2 + (b/a)x + (c/a)) = ax2 + bx + c.

Note that in some cases, the quadratic will not factor easily, and you may get complex (imaginary) numbers.

Let’s look at an example of how this works for a quadratic divisor.

### Example: Synthetic Division With A Quadratic Divisor

Let’s say we want to divide 20x3 + 55x2 + 25x – 10 by 5x2 + 15x + 10.

Following the 5-step process above:

• 1.) Factor the quadratic divisor into a product of a constant and a quadratic monic polynomial.  Here, we factor as: 5x2 + 15x + 10 = 5(x2 + 3x + 2).  [Note that this leaves us with a quadratic monic polynomial in parentheses – that is, the coefficient of x2 is 1].
• 2.) Factor the quadratic monic polynomial into a product of two linear monic polynomials.  Here, we factor as: 5(x2 + 3x + 2) = 5(x + 1)(x + 2).
• 3.) Use synthetic division to divide the dividend polynomial 20x3 + 55x2 + 25x – 10  by the linear monic x + 1.  This gives us a result of 20x2 + 35x – 10.
• 4.) Take the result from step 3 and divide it by x + 2, using synthetic division again.  This gives us  result of 20x – 5.
• 5.) Take the result from step 4 and divide by a.  Here, a = 5, so we get a result of 4x – 1.

We can verify this result if we use the Distributive Property for (5x2 + 15x + 10)(4x – 1) and combine like terms to get 20x3 + 55x2 + 25x – 10.

Below, we see the details of steps 3 and 4 (synthetic division) in this process:

## Conclusion

Now you know how to do synthetic division with a linear polynomial divisor whose lead coefficient is not 1.  You also know how to do synthetic division with a quadratic divisor by using synthetic division repeatedly.

I hope you found this article helpful.  If so, please share it with someone who can use the information.

~Jonathon