Chapter 3 of 7

All 7 chapters
  1. 1Domain, range and inverses, readunderstood
  2. 2The modulus function, readunderstood
  3. 3Dividing polynomials, readunderstood
  4. 4The factor theorem, readunderstood
  5. 5Partial fractions, readunderstood
  6. 6Pascal's triangle and choosing, readunderstood
  7. 7The binomial expansion, readunderstood

M5.1.3 Chapter 3

Dividing polynomials

Can I divide x³ + 2x² − 5x − 6 by x + 1 the way I divide 86 by 7?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Divide x³ + 2x² − 5x − 6 by x + 1
  • Find a remainder quickly with the remainder theorem

A polynomial in x is a sum of terms. Each term is a number times a power of x, and the power is a whole number: 0, 1, 2, 3, and so on. For example 4x³ − 2x² + x + 7 is a polynomial.

The number in front of a power is its coefficient. The highest power of x is the degree of the polynomial. The term with no x is the constant term. So 1/x and √x are not polynomials.

A table with the heads term, number and power. The rows are 4x³, 4 and 3; −2x², −2 and 2; x, 1 and 1; 7, 7 and 0.termnumberpower4x³43−2x²−22x11770
The polynomial 4x³ − 2x² + x + 7 has degree 3. Its constant term is 7, and the coefficient of x is 1.

With numbers, 17 ÷ 5 = 3 remainder 2 means 17 = 5 × 3 + 2. The remainder is smaller than 5. With polynomials we say the same thing, with the word "smaller" meaning "lower degree".

If f(x) is divided by d(x), with quotient q(x) and remainder r(x), then f(x) = d(x) × q(x) + r(x), and r(x) has a lower degree than d(x). Dividing by x − a gives a remainder with degree 0: just a number.

A long division. The divisor x + 2 stands left of a bracket. Under the bracket is x² + 5x + 7. Above it, the quotient x + 3. Under the dividend are the rows x² + 2x, then 3x + 7, then 3x + 6, and last the remainder 1.x + 2x²+ 5x+ 7xx²+ 2x3x+ 7+ 33x+ 61
x² + 5x + 7 divided by x + 2. Each power of x has its own column, so the terms stay in line.

Worked example

Divide x² + 5x + 7 by x + 2.

  1. Divide the first terms: x² ÷ x = x. Write x above the x column.
  2. Multiply the divisor by x: x(x + 2) = x² + 2x. Write it under the dividend.
  3. Subtract. The x² terms cancel, and 5x − 2x = 3x.
  4. Bring down the 7. The new line is 3x + 7.
  5. Divide the first terms: 3x ÷ x = 3. Write + 3 above.
  6. Multiply: 3(x + 2) = 3x + 6. Subtract: (3x + 7) − (3x + 6) = 1.
  7. The remainder 1 has degree 0, which is lower than the degree 1 of the divisor. Stop.

Answer: The quotient is x + 3 and the remainder is 1.

To check, multiply back: (x + 2)(x + 3) + 1 = x² + 5x + 6 + 1 = x² + 5x + 7. This is the dividend, so the division is right.

Now a cubic. The steps repeat once for each power. Watch each line build up in the picture below.

Draw it first

Divide x³ + 2x² − 5x − 6 by x + 1

  1. A long division built in three steps. The divisor x + 1 stands left of a bracket over x³ + 2x² − 5x − 6. Step 1 writes x² above, then the row x³ + x² and the line x² − 5x. Step 2 writes + x above, then the row x² + x and the line −6x − 6. Step 3 writes − 6 above, then the row −6x − 6 and the remainder 0.x + 1x³+ 2x²− 5x− 6x²x³+ x²x²− 5x

    1 Divide x³ by x to get x². Multiply x²(x + 1) = x³ + x², subtract, and bring down to get x² − 5x.

  2. A long division built in three steps. The divisor x + 1 stands left of a bracket over x³ + 2x² − 5x − 6. Step 1 writes x² above, then the row x³ + x² and the line x² − 5x. Step 2 writes + x above, then the row x² + x and the line −6x − 6. Step 3 writes − 6 above, then the row −6x − 6 and the remainder 0.x + 1x³+ 2x²− 5x− 6x²x³+ x²x²− 5x+ xx²+ x−6x− 6

    2 Divide x² by x to get x. Multiply x(x + 1) = x² + x, subtract, and bring down to get −6x − 6.

  3. A long division built in three steps. The divisor x + 1 stands left of a bracket over x³ + 2x² − 5x − 6. Step 1 writes x² above, then the row x³ + x² and the line x² − 5x. Step 2 writes + x above, then the row x² + x and the line −6x − 6. Step 3 writes − 6 above, then the row −6x − 6 and the remainder 0.x + 1x³+ 2x²− 5x− 6x²x³+ x²x²− 5x+ xx²+ x−6x− 6− 6−6x− 60

    3 Divide −6x by x to get −6. Multiply −6(x + 1) = −6x − 6, subtract, and the remainder is 0.

Worked example

Divide x³ + 2x² − 5x − 6 by x + 1.

  1. Divide the first terms: x³ ÷ x = x². Write x² above the x² column.
  2. Multiply: x²(x + 1) = x³ + x². Subtract from the top: (x³ + 2x²) − (x³ + x²) = x². Bring down −5x to get x² − 5x.
  3. Divide the first terms: x² ÷ x = x. Write + x above.
  4. Multiply: x(x + 1) = x² + x. Subtract: (x² − 5x) − (x² + x) = −6x. Bring down −6 to get −6x − 6.
  5. Divide the first terms. Write − 6 above, because −6x ÷ x = −6.
  6. Multiply: −6(x + 1) = −6x − 6. Subtract: (−6x − 6) − (−6x − 6) = 0.
  7. The remainder is 0, so x + 1 divides exactly. So x³ + 2x² − 5x − 6 = (x + 1)(x² + x − 6).

Answer: The quotient is x² + x − 6 and the remainder is 0.

If a power of x is missing, write it in with coefficient 0. This keeps the columns in line. So x³ − 7x + 6 is written x³ + 0x² − 7x + 6.

A long division. The divisor x − 1 stands left of a bracket over x³ + 0x² − 7x + 6. The quotient above is x² + x − 6. The last line is the remainder 0.x − 1x³+ 0x²− 7x+ 6x²x³− x²x²− 7x+ xx²− x−6x+ 6− 6−6x+ 60
x³ − 7x + 6 divided by x − 1. The missing x² term is written as + 0x², so that −7x stays in the x column.

Worked example

Divide x³ − 4x² + x + 9 by x − 2. Write the result as f(x) = (x − 2) × q(x) + r.

  1. Divide x³ ÷ x = x². Multiply: x²(x − 2) = x³ − 2x². Subtract: (x³ − 4x²) − (x³ − 2x²) = −2x². Bring down x: −2x² + x.
  2. Divide −2x² ÷ x = −2x. Multiply: −2x(x − 2) = −2x² + 4x. Subtract: (−2x² + x) − (−2x² + 4x) = −3x. Bring down 9: −3x + 9.
  3. Divide −3x ÷ x = −3. Multiply: −3(x − 2) = −3x + 6. Subtract: (−3x + 9) − (−3x + 6) = 3.
  4. The quotient is x² − 2x − 3 and the remainder is 3.
  5. So x³ − 4x² + x + 9 = (x − 2)(x² − 2x − 3) + 3.
  6. Check at x = 2. The bracket (x − 2) is 0, so the right side is 3. And the left side is 8 − 16 + 2 + 9 = 3.

Answer: The quotient is x² − 2x − 3 and the remainder is 3.

The check in the last step is the key. When x = 2 the divisor x − 2 is zero, so the whole product vanishes and only the remainder is left. The remainder equals the value of the polynomial at x = 2.

A graph with x from 0 to 4 and y from 0 to 10. A curve named y = f(x) falls from the point (0, 9) to a low point near x = 2.5, then rises. A dot with dashed guides marks the point (2, 3).(2, 3)y = f(x)12342468100xy
f(x) = x³ − 4x² + x + 9. At x = 2 the height of the curve is 3, the remainder when f(x) is divided by x − 2.

The remainder theorem says: when a polynomial f(x) is divided by x − a, the remainder is f(a).

Proof. Divide by x − a. The remainder has degree less than 1, so it is a number r. Then f(x) = (x − a) q(x) + r for every x. Put x = a. The bracket is 0, so f(a) = 0 × q(a) + r = r.

Take care with the sign. The divisor x + 2 is x − (−2), so a = −2 and the remainder is f(−2). In general, the divisor x + b gives f(−b), and the divisor 2x − 1 gives f(1/2).

Worked example

Find the remainder when f(x) = 2x³ − 5x² + x + 7 is divided by x + 2.

  1. Write the divisor as x − (−2). So a = −2.
  2. By the remainder theorem the remainder is f(−2).
  3. Work out each term: 2 × (−2)³ = 2 × (−8) = −16, and −5 × (−2)² = −5 × 4 = −20.
  4. The other terms: x = −2, and the constant is 7.
  5. Add: f(−2) = −16 − 20 − 2 + 7 = −31.

Answer: The remainder is −31.

Worked example

The polynomial f(x) = x³ + px² − 4x + 3 leaves remainder 11 when divided by x − 2. Find p. Then find the remainder when f(x) is divided by x + 1.

  1. The divisor is x − 2, so a = 2 and the remainder is f(2).
  2. f(2) = 2³ + p × 2² − 4 × 2 + 3 = 8 + 4p − 8 + 3 = 4p + 3.
  3. The remainder is 11, so 4p + 3 = 11. Then 4p = 8, so p = 2.
  4. So f(x) = x³ + 2x² − 4x + 3.
  5. The divisor x + 1 is x − (−1), so the remainder is f(−1).
  6. f(−1) = (−1)³ + 2 × (−1)² − 4 × (−1) + 3 = −1 + 2 + 4 + 3 = 8.

Answer: p = 2, and the second remainder is 8.

For a divisor such as 2x − 1, write f(x) = (2x − 1)q(x) + r. At x = 1/2 the bracket 2x − 1 is 0, so r = f(1/2). In general, the divisor px − q gives the remainder f(q/p), where q/p is the value of x that makes the divisor 0.

Worked example

Find the remainder when f(x) = 4x³ − 2x + 5 is divided by 2x − 1.

  1. The divisor 2x − 1 is 0 when x = 1/2. So the remainder is f(1/2).
  2. f(1/2) = 4 × (1/2)³ − 2 × (1/2) + 5.
  3. (1/2)³ = 1/8, so 4 × 1/8 = 1/2. And 2 × 1/2 = 1.
  4. So f(1/2) = 1/2 − 1 + 5 = 9/2.

Answer: The remainder is 9/2.

A divisor can have degree 2 as well. The remainder then has degree 1 or less, so it can hold an x term. The long division is the same.

A long division. The divisor x² + 1 stands left of a bracket over x³ + 3x² − 2x + 4. The quotient above is x + 3. Under the dividend are the rows x³ + x, then 3x² − 3x, then 3x² + 3, and last the remainder −3x + 1.x² + 1x³+ 3x²− 2x+ 4xx³+ x3x²− 3x+ 4+ 33x²+ 3−3x+ 1
x³ + 3x² − 2x + 4 divided by x² + 1. The quotient is x + 3. The remainder −3x + 1 has degree 1, which is less than 2.

Worked example

Divide x³ + 3x² − 2x + 4 by x² + 1.

  1. Divide the first terms: x³ ÷ x² = x. Write x above the x column.
  2. Multiply: x(x² + 1) = x³ + x. Subtract: (x³ + 3x²) − (x³ + 0x²) = 3x². The x column gives −2x − x = −3x. So the line is 3x² − 3x.
  3. Divide the first terms: 3x² ÷ x² = 3. Write + 3 above.
  4. Multiply: 3(x² + 1) = 3x² + 3. Subtract: (3x² − 3x + 4) − (3x² + 3) = −3x + 1.
  5. The remainder −3x + 1 has degree 1, which is less than the degree 2 of the divisor. Stop.

Answer: The quotient is x + 3 and the remainder is −3x + 1.

Try it

Multiply back to check a division

You need: paper, a pencil.

  1. Take the result of the first worked example: quotient x + 3, divisor x + 2, remainder 1.
  2. Draw a grid with x and + 2 across, and x and + 3 down. Multiply each pair of terms.
  3. Add the four products and collect like terms.
  4. Add the remainder 1. Compare with the dividend x² + 5x + 7.

What you will see: The grid gives x² + 5x + 6. Adding the remainder 1 gives x² + 5x + 7, the dividend. A division is right when dividend = divisor × quotient + remainder.

If you cannot try it, this is what happens.
A grid with x and +2 across the top and x and +3 down the side. The cells read x², 2x, 3x and 6. Under the grid is the sum = x² + 5x + 6.x+2x+3x²+2x+3x+6= x² + 5x + 6

A wrong idea many people have

To find the remainder when dividing by x + 2, work out f(2).

Why it feels right: The number 2 is the one written in the divisor, so it seems the one to put in.

The test that breaks it: Take f(x) = x + 2 itself. It divides exactly by x + 2, so the remainder is 0. But f(2) = 4, and f(−2) = 0.

What is true: The divisor x + 2 is x − (−2). Put in x = −2. Write the divisor as x − a first, and then the number is a.

A wrong idea many people have

A missing power of x can be left out of the long division.

Why it feels right: The term is not there, so it seems there is nothing to write.

The test that breaks it: In x³ − 7x + 6 the −7x would then stand in the x² column. The first subtraction would mix an x term with an x² term, and the answer would be wrong.

What is true: Write the missing power with coefficient 0: x³ + 0x² − 7x + 6. Each column then holds one power.

A wrong idea many people have

To subtract (x² − 5x) − (x² + x) you get −5x + x = −4x.

Why it feels right: The first term cancels, and then we carry on adding what is left.

The test that breaks it: Multiply back: (x + 1)(x² + x − 4) = x³ + 2x² − 3x − 4. This is not the dividend x³ + 2x² − 5x − 6, so the division is wrong.

What is true: Subtract every term of the line below: change the sign of each term. (x² − 5x) − (x² + x) = −5x − x = −6x.

Key words

polynomial
a sum of terms, each a number times a whole number power of x.
coefficient
the number in front of a power of x.
degree
the highest power of x in a polynomial.
quotient
the result of a division, without the remainder.
remainder
what is left when the division stops. It has a lower degree than the divisor.
remainder theorem
the remainder of f(x) divided by x − a is f(a).

Back to your guess

The answer is B: Yes. The same steps work, one power of x at a time.

The steps for long division of numbers work for polynomials too. At each step you divide the first term by the first term of the divisor, multiply back and subtract. A remainder is allowed, as it is with numbers, and it can be found without any division at all.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. What is the degree of the polynomial 5x⁴ − 3x² + 7?

    Tap your answer.

    1. Yes, that is right. The powers of x are 4, 2 and 0 (for the constant 7). The highest is 4.

    2. Not this one. You took the power of the second term. The degree is the highest power, wherever it stands. Go to M5.1.3. Try another.

    3. Not this one. You counted the terms. The degree is the highest power of x, not the number of terms. Go to M5.1.3. Try another.

    4. Not this one. That is a coefficient, the number in front of x⁴. The degree is the power, 4. Go to M5.1.3. Try another.

    See the answer

    4

    The powers of x are 4, 2 and 0 (for the constant 7). The highest is 4.

    • If you chose B: You took the power of the second term. The degree is the highest power, wherever it stands. Go to M5.1.3.
    • If you chose C: You counted the terms. The degree is the highest power of x, not the number of terms. Go to M5.1.3.
    • If you chose D: That is a coefficient, the number in front of x⁴. The degree is the power, 4. Go to M5.1.3.

  2. What is the remainder when x³ − 2x + 5 is divided by x − 3?

    Tap your answer.

    1. Not this one. You left out the +5. f(3) = 27 − 6 + 5, which needs every term. Go to M5.1.3. Try another.

    2. Not this one. You worked out f(−3). The divisor is x − 3, so a = 3 and the remainder is f(3). Go to M5.1.3. Try another.

    3. Not this one. That is f(0), the constant term. It is the remainder only when you divide by x. Go to M5.1.3. Try another.

    4. Yes, that is right. The divisor is x − 3, so a = 3. The remainder is f(3) = 3³ − 2 × 3 + 5 = 27 − 6 + 5 = 26.

    See the answer

    26

    The divisor is x − 3, so a = 3. The remainder is f(3) = 3³ − 2 × 3 + 5 = 27 − 6 + 5 = 26.

    • If you chose A: You left out the +5. f(3) = 27 − 6 + 5, which needs every term. Go to M5.1.3.
    • If you chose B: You worked out f(−3). The divisor is x − 3, so a = 3 and the remainder is f(3). Go to M5.1.3.
    • If you chose C: That is f(0), the constant term. It is the remainder only when you divide by x. Go to M5.1.3.

  3. x + 4 divides the polynomial f(x) exactly, with no remainder. Which statement must be true?

    Tap your answer.

    1. Yes, that is right. The remainder on dividing by x + 4 is f(−4). Dividing exactly means the remainder is 0, so f(−4) = 0.

    2. Not this one. The divisor x + 4 is x − (−4), so the number to put in is −4, not 4. Go to M5.1.3. Try another.

    3. Not this one. That mixes up the number in the divisor with the value of f. The remainder is f(−4), and here it is 0. Go to M5.1.3. Try another.

    4. Not this one. The remainder is f(−4), and "divides exactly" means the remainder is 0, not 4. Go to M5.1.3. Try another.

    See the answer

    f(−4) = 0

    The remainder on dividing by x + 4 is f(−4). Dividing exactly means the remainder is 0, so f(−4) = 0.

    • If you chose B: The divisor x + 4 is x − (−4), so the number to put in is −4, not 4. Go to M5.1.3.
    • If you chose C: That mixes up the number in the divisor with the value of f. The remainder is f(−4), and here it is 0. Go to M5.1.3.
    • If you chose D: The remainder is f(−4), and "divides exactly" means the remainder is 0, not 4. Go to M5.1.3.

  4. The picture shows the first step of dividing x³ + 4x² + x − 6 by x + 2. What is the next term of the quotient?

    A long division begun. The divisor x + 2 stands left of a bracket over x³ + 4x² + x − 6. The quotient so far is x². Under the dividend is the row x³ + 2x², and under it the line 2x² + x.x + 2x³+ 4x²+ x− 6x²x³+ 2x²2x²+ x
    Tap your answer.

    1. Not this one. That is the first term of the new line, not a quotient term. Divide it by x, the first term of the divisor. Go to M5.1.3. Try another.

    2. Not this one. You divided the numbers but forgot the powers. 2x² ÷ x = 2x, because x² ÷ x = x. Go to M5.1.3. Try another.

    3. Yes, that is right. Divide the first term of the line, 2x², by the first term of the divisor, x. That gives 2x.

    4. Not this one. That is the divisor. The quotient term is the first term of the line divided by the first term of the divisor. Go to M5.1.3. Try another.

    See the answer

    2x

    Divide the first term of the line, 2x², by the first term of the divisor, x. That gives 2x.

    • If you chose A: That is the first term of the new line, not a quotient term. Divide it by x, the first term of the divisor. Go to M5.1.3.
    • If you chose B: You divided the numbers but forgot the powers. 2x² ÷ x = 2x, because x² ÷ x = x. Go to M5.1.3.
    • If you chose D: That is the divisor. The quotient term is the first term of the line divided by the first term of the divisor. Go to M5.1.3.

  5. Amina divides f(x) = x² + 3x − 4 by x − 1 and says the remainder is f(−1) = −6. Which statement is right?

    Tap your answer.

    1. Not this one. The divisor x − 1 has a = 1. The sign in the divisor is the opposite of the number to put in, so the number is 1, not −1. Go to M5.1.3. Try another.

    2. Yes, that is right. The divisor x − 1 gives a = 1. The remainder is f(1) = 1 + 3 − 4 = 0, so x − 1 divides exactly.

    3. Not this one. The 4 comes from the constant term −4 of f. It does not come from the divisor. For x − 1 the number is 1. Go to M5.1.3. Try another.

    See the answer

    Use f(1), so the remainder is 0.

    The divisor x − 1 gives a = 1. The remainder is f(1) = 1 + 3 − 4 = 0, so x − 1 divides exactly.

    • If you chose A: The divisor x − 1 has a = 1. The sign in the divisor is the opposite of the number to put in, so the number is 1, not −1. Go to M5.1.3.
    • If you chose C: The 4 comes from the constant term −4 of f. It does not come from the divisor. For x − 1 the number is 1. Go to M5.1.3.

  6. Find the remainder when x³ − 2x² + 3x + 4 is divided by x − 2.

    See the answer

    10

    The divisor is x − 2, so a = 2. The remainder is f(2) = 8 − 8 + 6 + 4 = 10.

  7. The polynomial f(x) = x³ + kx² − 5x + 1 leaves remainder 11 when divided by x − 2. Find k.

    See the answer

    3

    The remainder is f(2) = 8 + 4k − 10 + 1 = 4k − 1. This equals 11, so 4k = 12 and k = 3.

  8. Divide x³ − 6x² + 11x − 5 by x − 1. Write the quotient and the remainder.

    See the answer

    The quotient is x² − 5x + 6 and the remainder is 1.

    x³ ÷ x = x². Multiply: x³ − x². Subtract: −5x² + 11x. Then −5x² ÷ x = −5x. Multiply: −5x² + 5x. Subtract: 6x − 5. Then 6x ÷ x = 6. Multiply: 6x − 6. Subtract: −5 − (−6) = 1. The remainder 1 equals f(1) = 1 − 6 + 11 − 5 = 1.

  9. From an earlier chapter

    Which values of x satisfy |x − 3| < 2?

    Tap your answer.

    1. Yes, that is right. The distance from x to 3 is less than 2, so x lies between 3 − 2 = 1 and 3 + 2 = 5.

    2. Not this one. You used only x − 3 < 2. Both limits count: x − 3 must also be more than −2. Go to M5.1.2. Try another.

    3. Not this one. That is the answer to |x − 3| > 2, the numbers far from 3. A modulus below a number means close to 3. Go to M5.1.2. Try another.

    See the answer

    1 < x < 5

    The distance from x to 3 is less than 2, so x lies between 3 − 2 = 1 and 3 + 2 = 5.

    • If you chose B: You used only x − 3 < 2. Both limits count: x − 3 must also be more than −2. Go to M5.1.2.
    • If you chose C: That is the answer to |x − 3| > 2, the numbers far from 3. A modulus below a number means close to 3. Go to M5.1.2.

  10. f(x) = x³ + ax² + bx − 6. The expression x − 2 is a factor of f(x), and the remainder is −12 when f(x) is divided by x + 1. Find b.

    See the answer

    3

    x − 2 is a factor, so f(2) = 0: 8 + 4a + 2b − 6 = 0, so 4a + 2b = −2, which is 2a + b = −1. The remainder on dividing by x + 1 is f(−1) = −12: −1 + a − b − 6 = −12, so a − b = −5. Add the two equations: 3a = −6, so a = −2. Then b = −1 − 2a = 3.

Keep it

Copy this into your exercise book.

Divide first terms, multiply back, subtract, bring down. dividend = divisor × quotient + remainder. Remainder of f(x) ÷ (x − a) is f(a). Divisor x + 2 means a = −2.

Draw: A long division in columns, one column for each power of x, with 0 as a place holder.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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