Polynomials and The Factor Theorem
This topic covers the core skills for manipulating polynomials, which are expressions like 3x³ - x + 4. Mastering these techniques—expanding, factorising, and dividing—is essential for solving higher-order equations and analysing function behaviour in the ESAT.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The Factor Theorem provides a direct link between the roots of a polynomial equation and its linear factors. If P(a) = 0, then (x - a) is a factor.
- The Remainder Theorem is a shortcut to find the remainder of a polynomial division. The remainder when P(x) is divided by (x - a) is simply the value P(a).
- Algebraic long division is a robust method to divide polynomials by linear or quadratic expressions. It is crucial for finding the quotient after a factor has been identified.
- A standard strategy for factorising a cubic is to use the Factor Theorem to find an integer root 'a', which gives a linear factor (x - a). Then, divide the cubic by this factor to obtain a quadratic, which can then be factorised further.
- When setting up algebraic long division, always include placeholder terms for missing powers of x (e.g., write x³ + 2x - 1 as x³ + 0x² + 2x - 1) to maintain correct column alignment.
Formulae
P(x) = (x - a)Q(x) + R This general identity represents polynomial division, where P(x) is the original polynomial, (x-a) is the divisor, Q(x) is the quotient, and R is the remainder.
P(a) = 0 ≤> (x - a) is a factor of P(x) This is the Factor Theorem. Use it to test potential roots to find factors, or use a known factor to establish a root.
When P(x) is divided by (x - a), the remainder is P(a) This is the Remainder Theorem. Use it to find a remainder quickly without performing the full division.
Definitions
- Polynomial
- An algebraic expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. For example, 5x⁴ - 2x² + 9.
- Root
- A value of the variable that makes a polynomial equal to zero. If 'a' is a root of P(x), then P(a) = 0. This is also called a 'zero' of the polynomial.
- Factor
- A polynomial that divides another polynomial exactly, with a remainder of zero.
Worked example
The polynomial P(x) = x³ + kx² + 7x - 10 has a factor of (x - 2). When P(x) is divided by (x + 3), what is the remainder?
- 1
First, use the Factor Theorem to find the value of k.
Since (x - 2) is a factor, we know that P(2) = 0.
- 2
Substitute x = 2 into the polynomial:
(2)³ + k(2)² + 7(2) - 10 = 0.
- 3
Simplify and solve for k:
8 + 4k + 14 - 10 = 0 ⇒ 4k + 12 = 0 ⇒ k = -3.
- 4
Now we have the full polynomial:
P(x) = x³ - 3x² + 7x - 10 - 5
To find the remainder when P(x) is divided by (x + 3), use the Remainder Theorem.
The remainder is P(-3).
- 6
Calculate P(-3):
(-3)³ - 3(-3)² + 7(-3) - 10 = -27 - 3(9) - 21 - 10.
- 7
Complete the arithmetic:
-27 - 27 - 21 - 10 = -85.
Answer: -85
Common mistakes
- ×Sign errors in the theorems: A common mistake is mixing up (x - a) and (x + a). Remember, if the factor is (x - a), you test P(a). If the divisor is (x + a), you find the remainder by calculating P(-a).
- ×Errors with negative numbers: When substituting a negative value like -2 into a polynomial, be careful with signs. For example, (-2)³ is -8, but -3(-2)² is -3(4) = -12. A slip here will derail the entire question.
- ×Subtraction in long division: When performing long division, you subtract an entire line of terms. A frequent error is mishandling the signs, e.g., (5x) - (-2x) becomes 3x instead of the correct 7x. Always think 'change the signs and add'.
No-calculator tips
- ✓Rational Root Theorem shortcut: To find an integer root for a polynomial, only test integer factors of the constant term. For x³ + 2x² - 5x - 6, you only need to test ±1, ±2, ±3, and ±6, not other numbers like 4 or 5.
- ✓Quickly check for factor (x - 1): To test if (x - 1) is a factor, simply add up all the coefficients of the polynomial. If the sum is zero, then (x - 1) is a factor. For x³ - 2x² - 5x + 6, the sum is 1 - 2 - 5 + 6 = 0, so (x-1) is a factor.
- ✓Factor check by inspection: Once you have one linear factor (e.g., x-2) and have divided to get a quadratic (e.g., ax²+bx+c), do a quick mental check. The constant term of the linear factor (-2) times the constant of the quadratic (c) must equal the constant of the original cubic.