Warm Up

Warm Up

Without a calculator divide the following problems using long division.

a 218 ÷ 7

b 5361 ÷ 9

c 2712 ÷ 26

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 1 / 20

Polynomial and Synthetic Division

Pre-Calculus

Mr. Niedert

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 2 / 20

Polynomial and Synthetic Division

1 Long Division of Polynomials

2 Synthetic Division

3 The Remainder Theorem

4 The Factor Theorem

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 3 / 20

Polynomial and Synthetic Division

1 Long Division of Polynomials

2 Synthetic Division

3 The Remainder Theorem

4 The Factor Theorem

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 3 / 20

Polynomial and Synthetic Division

1 Long Division of Polynomials

2 Synthetic Division

3 The Remainder Theorem

4 The Factor Theorem

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 3 / 20

Polynomial and Synthetic Division

1 Long Division of Polynomials

2 Synthetic Division

3 The Remainder Theorem

4 The Factor Theorem

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 3 / 20

Purpose of Polynomial Long Division

Suppose that you are given the graph off (x) = 6x3 − 19x2 + 16x − 4, as seen below.

We can see that f has a zero at x = 2, but it is unclear where theother two zeros exist.

If we are able to remove (x − 2) as a factor then we are able to workwith a quadratic function, as opposed to a cubic, and we know howto find the zeros of any quadratic function.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 4 / 20

Purpose of Polynomial Long Division

Suppose that you are given the graph off (x) = 6x3 − 19x2 + 16x − 4, as seen below.

We can see that f has a zero at x = 2, but it is unclear where theother two zeros exist.

If we are able to remove (x − 2) as a factor then we are able to workwith a quadratic function, as opposed to a cubic, and we know howto find the zeros of any quadratic function.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 4 / 20

Purpose of Polynomial Long Division

Suppose that you are given the graph off (x) = 6x3 − 19x2 + 16x − 4, as seen below.

We can see that f has a zero at x = 2, but it is unclear where theother two zeros exist.

If we are able to remove (x − 2) as a factor then we are able to workwith a quadratic function, as opposed to a cubic, and we know howto find the zeros of any quadratic function.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 4 / 20

Long Division of Polynomials

Example

Divide(6x3 − 19x2 + 16x − 4

)by (x − 2), and use the result to factor the

polynomial completely.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 5 / 20

Long Division of Polynomials

Practice

Divide the following polynomials using polynomial long division. Use theresult to factor the polynomial completely.

a(5x2 − 17x − 12

)÷ (x − 4)

b(x4 − 1

)÷ (x + 1)

c(6x4 − x3 − x2 + 9x − 3

)÷(x2 + x − 1

)

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 6 / 20

Polynomial and Synthetic Division (Part 1 of 3)Assignment

pg. 159-160 Exercises #5-15 odd

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 7 / 20

Purpose of Synthetic Division

When dividing by divisors of the form x − k , we can use syntheticdivision as a shortcut for polynomial long division.

Keep in mind though that synthetic division works only for divisor ofthe form x − k . Specifically, you cannot use synthetic division todivide a polynomial by a quadratic polynomial (such as x2 + 2x + 1)or any other higher degree polynomial.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 8 / 20

Purpose of Synthetic Division

When dividing by divisors of the form x − k , we can use syntheticdivision as a shortcut for polynomial long division.

Keep in mind though that synthetic division works only for divisor ofthe form x − k . Specifically, you cannot use synthetic division todivide a polynomial by a quadratic polynomial (such as x2 + 2x + 1)or any other higher degree polynomial.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 8 / 20

Using Synthetic Division

Example

Use synthetic division to divide x4 − 10x2 − 2x + 4 by x + 3.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 9 / 20

Using Synthetic Division

Practice

Use synthetic division to divide 5x3 + 8x2 − x + 6 by x + 2.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 10 / 20

Two Common Errors with Synthetic Long Division toAvoid

1 Missing Powers of x

I If there are missing powers of x (as in the first example today), theremust be a placeholder in the synthetic division for each missing term.

I For example, x5 − 32 would be represented as 1 0 0 0 0 − 32.

2 Using AdditionI When doing long division we subtract the columns, but when using

synthetic division we add the columns.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 11 / 20

Two Common Errors with Synthetic Long Division toAvoid

1 Missing Powers of xI If there are missing powers of x (as in the first example today), there

must be a placeholder in the synthetic division for each missing term.

I For example, x5 − 32 would be represented as 1 0 0 0 0 − 32.

2 Using AdditionI When doing long division we subtract the columns, but when using

synthetic division we add the columns.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 11 / 20

Two Common Errors with Synthetic Long Division toAvoid

1 Missing Powers of xI If there are missing powers of x (as in the first example today), there

must be a placeholder in the synthetic division for each missing term.I For example, x5 − 32 would be represented as 1 0 0 0 0 − 32.

2 Using AdditionI When doing long division we subtract the columns, but when using

synthetic division we add the columns.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 11 / 20

Two Common Errors with Synthetic Long Division toAvoid

1 Missing Powers of xI If there are missing powers of x (as in the first example today), there

must be a placeholder in the synthetic division for each missing term.I For example, x5 − 32 would be represented as 1 0 0 0 0 − 32.

2 Using Addition

I When doing long division we subtract the columns, but when usingsynthetic division we add the columns.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 11 / 20

Two Common Errors with Synthetic Long Division toAvoid

1 Missing Powers of xI If there are missing powers of x (as in the first example today), there

must be a placeholder in the synthetic division for each missing term.I For example, x5 − 32 would be represented as 1 0 0 0 0 − 32.

2 Using AdditionI When doing long division we subtract the columns, but when using

synthetic division we add the columns.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 11 / 20

Polynomial and Synthetic Division (Part 2 of 3)Assignment

pg. 159-160 Exercises #5-15 odd, 19-35 EOO

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 12 / 20

The Remainder Theorem

Theorem

If a polynomial f (x) is divided by x − k , the remainder is r = f (k).

The Remainder Theorem tells us that the synthetic division can beused to evaluate a polynomial function.

That is, to evaluate a polynomial function f (x) when x = k , dividef (x) by x − k .

The remainder will be f (k).

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 13 / 20

The Remainder Theorem

Theorem

If a polynomial f (x) is divided by x − k , the remainder is r = f (k).

The Remainder Theorem tells us that the synthetic division can beused to evaluate a polynomial function.

That is, to evaluate a polynomial function f (x) when x = k , dividef (x) by x − k .

The remainder will be f (k).

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 13 / 20

The Remainder Theorem

Theorem

If a polynomial f (x) is divided by x − k , the remainder is r = f (k).

The Remainder Theorem tells us that the synthetic division can beused to evaluate a polynomial function.

That is, to evaluate a polynomial function f (x) when x = k , dividef (x) by x − k .

The remainder will be f (k).

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 13 / 20

Using the Remainder Theorem

Example

Use the Remainder Theorem to evaluate the functionf (x) = 3x3 + 8x2 + 5x − 7 at x = −2.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 14 / 20

Using the Remainder Theorem

Practice

Use the Remainder Theorem to evaluate the functionf (x) = 4x3 + 10x2 − 3x − 8 at the value f (−1).

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 15 / 20

The Factor Theorem

Theorem

A polynomial f (x) has a factor (x − k) if and only if f (k) = 0.

This theorem states that you can test to see whether a polynomialhas (x − k) as a factor by evaluating the polynomial at x = k.

If the result is 0, (x − k) is a factor.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 16 / 20

The Factor Theorem

Theorem

A polynomial f (x) has a factor (x − k) if and only if f (k) = 0.

This theorem states that you can test to see whether a polynomialhas (x − k) as a factor by evaluating the polynomial at x = k.

If the result is 0, (x − k) is a factor.

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 16 / 20

Factoring a Polynomial: Repeated Division

Example

Show that (x − 2) and (x + 3) of f (x) = 2x4 + 7x3 − 4x2 − 27x − 18.Then find the remaining factors of f (x).

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 17 / 20

Factoring a Polynomial: Repeated Division

Practice

Show that (x + 3) is a factor of x3 − 19x − 30 = 0. Then find theremaining factors of f (x).

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 18 / 20

Uses of the Remainder of Synthetic Division

Uses of the Remainder of Synthetic Division

The remainder r , obtained in the synthetic division of f (x) by x − k ,provided the following information.

1 The remainder r gives the value of f at x = k . That is, r = f (k).

2 If r = 0, (x − k) is a factor of f (x).

3 If r = 0, (k, 0) is an x-intercept of the graph of f .

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 19 / 20

Uses of the Remainder of Synthetic Division

Uses of the Remainder of Synthetic Division

The remainder r , obtained in the synthetic division of f (x) by x − k ,provided the following information.

1 The remainder r gives the value of f at x = k . That is, r = f (k).

2 If r = 0, (x − k) is a factor of f (x).

3 If r = 0, (k, 0) is an x-intercept of the graph of f .

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 19 / 20

Uses of the Remainder of Synthetic Division

Uses of the Remainder of Synthetic Division

The remainder r , obtained in the synthetic division of f (x) by x − k ,provided the following information.

1 The remainder r gives the value of f at x = k . That is, r = f (k).

2 If r = 0, (x − k) is a factor of f (x).

3 If r = 0, (k, 0) is an x-intercept of the graph of f .

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 19 / 20

Uses of the Remainder of Synthetic Division

Uses of the Remainder of Synthetic Division

The remainder r , obtained in the synthetic division of f (x) by x − k ,provided the following information.

1 The remainder r gives the value of f at x = k . That is, r = f (k).

2 If r = 0, (x − k) is a factor of f (x).

3 If r = 0, (k, 0) is an x-intercept of the graph of f .

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 19 / 20

Polynomial and Synthetic Division (Part 3 of 3)Assignment

pg. 159-160 Exercises #5-15 odd, 19-35 EOO, 45-51 odd, 57-61 odd

Pre-Calculus Polynomial and Synthetic Division Mr. Niedert 20 / 20