Hello, and welcome to today's mathematics lesson. We are going to explore factorization of polynomials, focusing on two incredibly powerful tools — the Remainder Theorem and the Factor Theorem. By the end of this lesson, you will understand how to find remainders without long division, and how to break down complex polynomials into simpler factors.
Let us begin with some essential groundwork. A polynomial is an algebraic expression where variables have whole number powers. Think of expressions like ax + b or ax² + bx + c, or even higher degree polynomials like 3x⁴ – 2x³ + 5x – 1. The highest power of the variable tells us the degree of the polynomial.
When we write f(x) = 2x² – 5x – 7, we call this a function of x. The value of f(x) changes as x changes. For instance, if x equals 3, then f(3) = 2(3)² – 5(3) – 7 = 18 – 15 – 7 = –4. And if x equals negative 3, f(–3) = 2(–3)² – 5(–3) – 7 = 18 + 15 – 7 = 26. This idea of substituting values will be crucial for what follows.
Now, here is a fascinating observation that leads us to the Remainder Theorem. Suppose we divide x² – 5x + 8 by x – 2. The remainder turns out to be 2. But notice what happens when we simply substitute x equals 2 into the original polynomial: 2² – 5(2) + 8 = 4 – 10 + 8 = 2. The remainder equals f(2)!
Let us verify with another example. Divide 6x³ – 3x² + 8x – 5 by x + 3. The remainder is negative 218. Now compute f(–3): 6(–3)³ – 3(–3)² + 8(–3) – 5 = –162 – 27 – 24 – 5 = –218. Again, the remainder equals the function value!
This brings us to the formal statement of the Remainder Theorem. If f(x), a polynomial in x, is divided by (x – a), then the remainder equals f(a).
To apply this theorem, follow these steps. First, set the divisor equal to zero and solve for the variable. Second, substitute this value into the given polynomial and simplify. The result is your remainder — no long division needed!
Here is how this works in practice. Suppose when 2x³ – kx² + (5k – 3)x – 8 is divided by x – 2, the remainder is 14. We need to find k. Setting x – 2 = 0, we get x = 2. Substituting: 2(2)³ – k(2)² + (5k – 3)(2) – 8 = 14. This simplifies to 16 – 4k + 10k – 6 – 8 = 14, so 6k – 2 = 14, so 6k = 12, and therefore k = 2.
Another useful application: when two polynomials leave the same remainder upon division by the same linear factor. If 3x³ – ax² + 5x – 13 and (a + 1)x² – 7x + 5 both give the same remainder when divided by x – 3, we set their values at x equals 3 equal to each other. This yields 81 – 9a + 15 – 13 = 9a + 9 – 21 + 5, which simplifies to 83 – 9a = 9a – 7, so 18a = 90, so a equals 5.
Sometimes we encounter conditions involving multiple remainders. Consider x³ + ax² – bx – 8 which gives remainder zero when divided by x – 2, and remainder negative 30 when divided by x + 1. From the first condition: 2³ + a(2)² – b(2) – 8 = 0, which gives 8 + 4a – 2b – 8 = 0, so 4a – 2b = 0, thus 2a – b = 0 as equation one. From the second: (–1)³ + a(–1)² – b(–1) – 8 = –30, which gives –1 + a + b – 8 = –30, so a + b – 9 = –30, thus a + b = –21 as equation two. Solving these simultaneous equations, we find a = –7 and b = –14.
We can also determine what constant must be added or subtracted to achieve a specific remainder. If we want 2x³ – 3x² + x to give remainder 3 when divided by x – 2, we let k be the number added. Then 2(2)³ – 3(2)² + 2 + k = 3, giving 16 – 12 + 2 + k = 3, so 6 + k = 3, thus k = –3.
Now we arrive at an even more powerful result — the Factor Theorem. This is really a special case of the Remainder Theorem. When a polynomial f(x) is divided by x – a, the remainder is f(a). If this remainder equals zero, then x – a is a factor of f(x). Conversely, if f(a) = 0, then x – a is a factor.
For example, take f(x) = x² – 5x + 6. Computing f(3): 3² – 5(3) + 6 = 9 – 15 + 6 = 0. Since the remainder is zero, x – 3 is indeed a factor. You can verify: x² – 5x + 6 = (x – 3)(x – 2).
To test whether x – 1 is a factor of x⁶ – x⁵ + x⁴ + x³ – x² – x + 1, we evaluate at x equals 1: 1⁶ – 1⁵ + 1⁴ + 1³ – 1² – 1 + 1 = 1. Since this is not zero, x – 1 is not a factor.
When x – 2 is given as a factor of x² – 7x + 2a, we know f(2) = 0, so 2² – 7(2) + 2a = 0. This gives 4 – 14 + 2a = 0, so 2a = 10, yielding a = 5.
For a more involved case, if x – 2 is a factor of x³ + 2x² – kx + 10, then 2³ + 2(2)² – k(2) + 10 = 0, giving 8 + 8 – 2k + 10 = 0, so 26 – 2k = 0, thus k = 13. With this value, our polynomial becomes x³ + 2x² – 13x + 10. To check if x + 5 is also a factor, we test f(–5): (–5)³ + 2(–5)² – 13(–5) + 10 = –125 + 50 + 65 + 10 = 0. Yes, it is a factor!
When multiple factors are given, we obtain multiple equations. If x + 2 and x – 3 are both factors of x³ + ax + b, then f(–2) = 0 gives (–2)³ + a(–2) + b = 0, so –8 – 2a + b = 0, and f(3) = 0 gives 3³ + a(3) + b = 0, so 27 + 3a + b = 0. Solving: from the first, b = 8 + 2a. Substituting: 27 + 3a + 8 + 2a = 0, so 5a = –35, thus a = –7, and b = –6.
The true power of the Factor Theorem emerges when we use it to completely factorise polynomials. Once we find one factor using the theorem, we can divide to find the remaining quadratic factor, which we then factorise further.
Consider showing that 2x + 7 is a factor of 2x³ + 5x² – 11x – 14. Set 2x + 7 = 0, so x = –7/2. Evaluating: 2(–7/2)³ + 5(–7/2)² – 11(–7/2) – 14 equals –343/4 + 245/4 + 77/2 – 14 = –343/4 + 245/4 + 154/4 – 56/4 = 0. Confirmed!
Now we divide 2x³ + 5x² – 11x – 14 by 2x + 7 to get x² – x – 2. This further factorises as (x – 2)(x + 1). Thus the complete factorisation is (2x + 7)(x – 2)(x + 1).
Another example: factorise 2x³ + x² – 2x – 1 completely. Testing x equals 1: 2(1)³ + (1)² – 2(1) – 1 = 0. So x – 1 is a factor. Dividing, we obtain 2x² + 3x + 1, which factorises as 2x² + 2x + x + 1 = 2x(x+1) + 1(x+1) = (2x+1)(x+1). Complete factorisation: (x – 1)(x + 1)(2x + 1).
For a polynomial with unknown coefficients, suppose x³ + ax² + bx – 45 has factors x – 1 and x + 5. From f(1) = 0: 1³ + a(1)² + b(1) – 45 = 0, so 1 + a + b – 45 = 0, giving a + b = 44. From f(–5) = 0: (–5)³ + a(–5)² + b(–5) – 45 = 0, so –125 + 25a – 5b – 45 = 0, giving 25a – 5b = 170, thus 5a – b = 34. Solving: adding the two equations a + b = 44 and 5a – b = 34 gives 6a = 78, so a = 13, and b = 31.
The polynomial becomes x³ + 13x² + 31x – 45. Dividing by x – 1 gives x² + 14x + 45, which factorises as (x + 9)(x + 5). Final answer: (x – 1)(x + 5)(x + 9).
One final illustration: if x – 2 is a factor of 2x³ – x² – px – 2, find p and factorise completely. From f(2) = 0: 2(2)³ – (2)² – p(2) – 2 = 0, so 16 – 4 – 2p – 2 = 0, giving 10 – 2p = 0, thus p = 5. The polynomial is 2x³ – x² – 5x – 2. Dividing by x – 2 yields 2x² + 3x + 1, which factorises as 2x² + 2x + x + 1 = 2x(x + 1) + 1(x + 1) = (2x + 1)(x + 1). Complete factorisation: (x – 2)(x + 1)(2x + 1).
Let us recap the essential points from today's lesson.
First, the Remainder Theorem: when a polynomial f(x) is divided by x – a, the remainder equals f(a). This eliminates the need for long division to find remainders.
Second, the Factor Theorem: if f(a) = 0, then x – a is a factor of f(x), and conversely. This is the key to identifying factors without guessing.
Third, to find unknown coefficients, set up equations using the conditions that remainders equal specified values or equal zero for factors.
Fourth, complete factorisation involves finding one factor by the Factor Theorem, dividing to reduce the degree, then factorising the quotient.
Fifth, always verify your factors by substitution or by expanding your final answer.
Sixth, for divisors like x + a or 2x + b, solve for x and substitute that value — negative a, or negative b over 2 — into the polynomial.
You now possess powerful tools that transform complex polynomial problems into straightforward substitution exercises. The Remainder and Factor Theorems bridge the gap between algebra and function evaluation, revealing elegant structure within polynomials. Practice applying these theorems systematically, and you will find factorisation becoming increasingly intuitive. Until next time, keep exploring, keep questioning, and enjoy the beauty of mathematics.