What is a polynomial?
A polynomial in x is an expression made of terms of the form a·xⁿ, where a is a real number (the coefficient) and n is a whole number (0, 1, 2, …).
3x² − 5x + 2, y³ + 4 and 7 are polynomials.
1/x + 2 is not a polynomial, because 1/x = x⁻¹ has a negative exponent.
√x + 1 is not a polynomial, because √x = x^(1/2) has a fractional exponent.
Degree and types
The degree is the highest power of the variable with a non-zero coefficient.
By degree | Example | By number of terms | Example |
|---|---|---|---|
Constant (degree 0) | 7 | Monomial (1 term) | 4x³ |
Linear (degree 1) | 2x + 1 | Binomial (2 terms) | x² − 9 |
Quadratic (degree 2) | x² − 3x + 2 | Trinomial (3 terms) | x² + x + 1 |
Cubic (degree 3) | x³ − 1 |
The zero polynomial, 0, has no defined degree.
Value and zeroes of a polynomial
The value of p(x) at x = a is p(a), found by substituting a for x.
If p(a) = 0, then a is a zero of the polynomial.
Example. For p(x) = x² − 5x + 6: p(2) = 4 − 10 + 6 = 0 and p(3) = 9 − 15 + 6 = 0. So 2 and 3 are the zeroes.
Facts to remember:
A linear polynomial ax + b (a ≠ 0) has exactly one zero: x = −b/a.
A polynomial of degree n has at most n zeroes.
A zero can itself be 0. For example, 0 is a zero of x² − 2x.
The remainder theorem
When p(x) is divided by (x − a), the remainder is p(a).
This lets you find a remainder without doing long division.
Example. Find the remainder when p(x) = x³ + 3x² − 5x + 4 is divided by (x − 1).
p(1) = 1 + 3 − 5 + 4 = 3.
For division by (x + 2), use a = −2, since x + 2 = x − (−2).
The factor theorem
(x − a) is a factor of p(x) if and only if p(a) = 0.
Example. Is (x + 1) a factor of x³ + x² + x + 1?
p(−1) = −1 + 1 − 1 + 1 = 0. Yes, it is a factor.
Factorising using the factor theorem
Example. Factorise x³ − 6x² + 11x − 6.
Try small integers that divide the constant term 6. p(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor.
Divide: x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6).
Factorise the quadratic: x² − 5x + 6 = (x − 2)(x − 3).
So x³ − 6x² + 11x − 6 = (x − 1)(x − 2)(x − 3).
For a quadratic ax² + bx + c, split the middle term: find two numbers whose product is ac and whose sum is b.
Algebraic identities
Identity |
|---|
(x + y)² = x² + 2xy + y² |
(x − y)² = x² − 2xy + y² |
x² − y² = (x + y)(x − y) |
(x + a)(x + b) = x² + (a + b)x + ab |
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx |
(x + y)³ = x³ + y³ + 3xy(x + y) |
(x − y)³ = x³ − y³ − 3xy(x − y) |
x³ + y³ = (x + y)(x² − xy + y²) |
x³ − y³ = (x − y)(x² + xy + y²) |
x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx) |
A very useful special case: if x + y + z = 0, then x³ + y³ + z³ = 3xyz.
Example. Evaluate 104 × 96 without multiplying directly.
104 × 96 = (100 + 4)(100 − 4) = 100² − 4² = 10 000 − 16 = 9984.