Polynomials
Senior 2 · Mathematics
Polynomials
Lesson 2 of 5
25 min
Multiplying polynomials is how you find the area of a plot whose sides are expressions, or how a business multiplies price by quantity when both depend on the same variable. Dividing polynomials is how you break a polynomial into simpler factors, which is the key to solving equations of degree 3 and higher.
You will learn two ways to divide: long division, which works for any divisor, and synthetic division, a fast shortcut when the divisor is x − c.
Multiply every term of the first polynomial by every term of the second (the distributive law), then collect like terms.
Worked example 1
3x2(2x2 − 5x + 1) = 6x4 − 15x3 + 3x2.
Worked example 2
(2x + 3)(x2 − 4x + 5)
2x × each term: 2x3 − 8x2 + 10x. Then 3 × each term: 3x2 − 12x + 15. Collect: 2x3 − 5x2 − 2x + 15.
Check the degree: degree 1 times degree 2 gives degree 3 ✓.
Worked example 3
Divide 2x3 − 3x2 + 4x − 5 by x − 2.
Quotient 2x2 + x + 6, remainder 7. So 2x3 − 3x2 + 4x − 5 = (x − 2)(2x2 + x + 6) + 7. Expanding the right side confirms it.
If a power is missing, write it with a zero coefficient first: divide x3 − 8 as x3 + 0x2 + 0x − 8. By x − 2 this gives x2 + 2x + 4 with remainder 0.
Write only the coefficients and the number c. Bring the first coefficient down; then repeatedly multiply by c and add to the next coefficient.
Worked example 4
The same division, (2x3 − 3x2 + 4x − 5) ÷ (x − 2), so c = 2.
| c = 2 | 2 | −3 | 4 | −5 |
| 4 | 2 | 12 | ||
| 2 | 1 | 6 | 7 |
Bring down 2. 2 × 2 = 4, and −3 + 4 = 1. 1 × 2 = 2, and 4 + 2 = 6. 6 × 2 = 12, and −5 + 12 = 7. The bottom row gives the quotient 2x2 + x + 6 and the last number, 7, is the remainder, exactly as in long division.
Worked example 5
(x3 + 2x2 − 5x − 6) ÷ (x + 3). Since x + 3 = x − (−3), use c = −3.
| c = −3 | 1 | 2 | −5 | −6 |
| −3 | 3 | 6 | ||
| 1 | −1 | −2 | 0 |
Quotient x2 − x − 2, remainder 0, so x + 3 is a factor.
When P(x) is divided by x − c, the remainder equals P(c). Check with example 3: P(2) = 16 − 12 + 8 − 5 = 7 ✓. This gives a quick way to find a remainder without dividing at all.
Tip
Exam technique. In synthetic division, c is the number that makes the divisor zero: for x + 3 use −3, not 3. And never skip a missing power: put a 0 in its place, or every later column will be wrong.
Expand (x − 5)(2x + 1).
Solution: 2x2 + x − 10x − 5 = 2x2 − 9x − 5.
Expand (x + 2)(x2 − 3x + 4).
Solution: x3 − 3x2 + 4x + 2x2 − 6x + 8 = x3 − x2 − 2x + 8.
Use synthetic division to divide x3 − 4x2 + x + 6 by x − 2.
Solution: Coefficients 1, −4, 1, 6 with c = 2. Bring down 1; 1 × 2 = 2, −4 + 2 = −2; −2 × 2 = −4, 1 − 4 = −3; −3 × 2 = −6, 6 − 6 = 0. Quotient x2 − 2x − 3, remainder 0.
Find the remainder when x3 + 2x2 − 3x + 5 is divided by x − 1.
Solution: By the remainder theorem, P(1) = 1 + 2 − 3 + 5 = 5.
Divide x3 − 8 by x − 2.
Solution: Use coefficients 1, 0, 0, −8 and c = 2: 1; 0 + 2 = 2; 0 + 4 = 4; −8 + 8 = 0. Quotient x2 + 2x + 4, remainder 0.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
