Welcome to Factors of Polynomials!

Hello Mathletes! This chapter might look intimidating because of the big words like "Polynomials" and "Theorems," but don't worry. This is one of the most powerful and systematic sections in Additional Mathematics (0606).

We are essentially learning how to undo multiplication for expressions far more complex than simple quadratics. Mastering these techniques—the Remainder Theorem and the Factor Theorem—allows you to solve cubic equations and factor complicated expressions efficiently. Let's dive in!

(P.S. Since calculators are sometimes disallowed in Paper 1, practicing these algebraic skills is essential!)

1. Quick Review: What is a Polynomial?

A polynomial is just a mathematical expression made up of terms involving positive integer powers of a variable (usually \(x\)), like \(x^3 + 4x^2 - 7\). We usually write a polynomial using the notation \(P(x)\) or \(f(x)\).

  • A linear polynomial has degree 1 (e.g., \(x+5\)).
  • A quadratic polynomial has degree 2 (e.g., \(2x^2 - 3x + 1\)).
  • A cubic polynomial has degree 3 (e.g., \(x^3 + 2x - 10\)). This chapter mostly focuses on cubic polynomials.

When we talk about factors, we mean expressions that divide the polynomial perfectly, leaving zero remainder. For example, \((x-1)\) is a factor of \(x^2 - 1\).

2. The Remainder Theorem

3.1 Know and use the remainder theorem

The Remainder Theorem is a massive shortcut! Instead of doing complicated algebraic long division just to find the remainder, we can use simple substitution.

The Analogy: A Simple Division

Imagine you divide 17 by 5. You get 3, remainder 2. In polynomial terms, we are interested in that remainder.

The Rule:

If a polynomial, \(P(x)\), is divided by a linear expression \((x - a)\), the remainder is simply \(P(a)\).

This works because when you substitute \(x=a\) into the divisor \((x-a)\), the divisor becomes zero, leaving only the remainder behind.

Important Accessibility Tip: The Sign Flip!
If you are dividing by:

  • \((x - 3)\), you substitute \(x = +3\).
  • \((x + 2)\), you substitute \(x = -2\). (Since \(x + 2\) is the same as \(x - (-2)\)).

Step-by-Step Example

Find the remainder when \(P(x) = x^3 - 3x^2 + 5x + 1\) is divided by \((x - 2)\).

1. Identify \(a\): Since the divisor is \((x - 2)\), we use \(a = 2\).
2. Substitute \(a\) into \(P(x)\): The remainder \(R\) is \(P(2)\).
\[R = P(2) = (2)^3 - 3(2)^2 + 5(2) + 1\] 3. Calculate:
\[R = 8 - 3(4) + 10 + 1\] \[R = 8 - 12 + 11\] \[R = 7\]

The remainder is 7. You found it without any long division!

Quick Review: Remainder Theorem

Dividing \(P(x)\) by \((x - a)\) gives remainder \(P(a)\).

3. The Factor Theorem

3.1 Know and use the factor theorem

The Factor Theorem is simply the Remainder Theorem when the remainder is zero.

The Rule:

A linear expression \((x - a)\) is a factor of the polynomial \(P(x)\) if and only if \(P(a) = 0\).

Analogy: If you divide 10 by 5, the remainder is 0. That tells you that 5 is a factor of 10. The same logic applies here!

Memory Aid: F-A-C-T-O-R means F(a)=0.

Step-by-Step Example

Show that \((x + 1)\) is a factor of \(P(x) = x^3 + 2x^2 - x - 2\).

1. Identify \(a\): Since the factor is \((x + 1)\), we substitute \(x = -1\).
2. Substitute into \(P(x)\):
\[P(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2\] 3. Calculate:
\[P(-1) = -1 + 2(1) + 1 - 2\] \[P(-1) = -1 + 2 + 1 - 2\] \[P(-1) = 0\]

Since \(P(-1) = 0\), we have successfully shown that \((x + 1)\) is a factor of \(P(x)\).

4. Finding Factors of Polynomials and Solving Cubic Equations

3.2 Find factors of polynomials & 3.3 Solve cubic equations

The main goal of this chapter is usually to factor a cubic polynomial completely, meaning turning \(P(x)\) into \((x-a)(x-b)(x-c)\).

To solve a cubic equation \(P(x) = 0\), you must first find the linear factor using the Factor Theorem, and then find the resulting quadratic factor.

Step 1: Finding the First Linear Factor (Trial and Error)

We need to guess a value of \(a\) such that \(P(a) = 0\). Where do we get our guesses?

Trick for finding test values: Look at the constant term of the polynomial. The possible integer factors of the polynomial MUST correspond to the integer factors (positive and negative) of the constant term.

Example: For \(P(x) = x^3 - 4x^2 + x + 6\), the constant term is 6.
Possible integer values for \(a\) are the factors of 6: \(\pm 1, \pm 2, \pm 3, \pm 6\).

Start testing the easiest values first: \(+1, -1, +2, -2\).

  • Test \(x=1\): \(P(1) = (1)^3 - 4(1)^2 + 1 + 6 = 1 - 4 + 1 + 6 = 4\). (No luck).
  • Test \(x=-1\): \(P(-1) = (-1)^3 - 4(-1)^2 + (-1) + 6 = -1 - 4 - 1 + 6 = 0\). (Success!)

Since \(P(-1) = 0\), then \((x - (-1))\), or \((x + 1)\), is our first factor.

Step 2: Finding the Quadratic Factor (The Division)

Now we know \(x^3 - 4x^2 + x + 6 = (x + 1) \times Q(x)\), where \(Q(x)\) is a quadratic expression \((Ax^2 + Bx + C)\).

You must find \(Q(x)\) using either Algebraic Long Division or the method of Comparing Coefficients.

Method A: Algebraic Long Division (Essential Technique)

Don't worry if this seems tricky at first—it’s just like primary school division, but with \(x\)'s!

We divide \(x^3 - 4x^2 + x + 6\) by \((x + 1)\).

1. Divide the highest powers: \(x^3 / x = x^2\). Write \(x^2\) in the quotient.
2. Multiply: \(x^2(x + 1) = x^3 + x^2\).
3. Subtract: \((x^3 - 4x^2) - (x^3 + x^2) = -5x^2\). Bring down the next term (\(+x\)).
4. Repeat: Divide the new highest power: \(-5x^2 / x = -5x\). Write \(-5x\) in the quotient.
5. Multiply: \(-5x(x + 1) = -5x^2 - 5x\).
6. Subtract: \(( -5x^2 + x) - (-5x^2 - 5x) = 6x\). Bring down the last term (\(+6\)).
7. Repeat: Divide: \(6x / x = +6\). Write \(+6\) in the quotient.
8. Multiply: \(6(x + 1) = 6x + 6\).
9. Subtract: \((6x + 6) - (6x + 6) = 0\). (Remainder is zero, confirming it’s a factor!)

Thus, the quadratic factor is \(Q(x) = x^2 - 5x + 6\).

Method B: Comparing Coefficients (Observation)

We know: \(x^3 - 4x^2 + x + 6 = (x + 1)(Ax^2 + Bx + C)\)

  • Look at the \(x^3\) term: \(x \times Ax^2 = x^3\). Since A must be 1, \(A=1\).
  • Look at the Constant term: \(1 \times C = 6\). Therefore, \(C=6\).

    Now we have: \((x + 1)(x^2 + Bx + 6)\)

  • Look at the \(x^2\) term: The coefficient of \(x^2\) is \(-4\). The \(x^2\) terms come from: \((x)(Bx)\) + \((1)(x^2)\).
    So: \(B + 1 = -4\). Thus, \(B = -5\).

The quadratic factor is \(x^2 - 5x + 6\). (Check the \(x\) term: \(6x + Bx = 6x - 5x = x\). It matches!)

Step 3: Solve the Cubic Equation (Final Factorization)

To solve \(x^3 - 4x^2 + x + 6 = 0\), we use the factored form:
\[(x + 1)(x^2 - 5x + 6) = 0\]

1. Solve the linear factor: \(x + 1 = 0 \implies x = -1\).

2. Solve the quadratic factor: \(x^2 - 5x + 6 = 0\). (Factor this quadratic easily.)
\[(x - 2)(x - 3) = 0\]
This gives \(x = 2\) and \(x = 3\).

The solutions to the cubic equation are \(x = -1, x = 2, x = 3\).

Key Takeaway: Solving Cubics

1. Use Factor Theorem (trial and error based on the constant term) to find the first linear factor \((x-a)\).

2. Use Algebraic Long Division or Comparing Coefficients to find the quadratic factor \(Q(x)\).

3. Solve \((x-a)=0\) and \(Q(x)=0\) to find all three roots.

5. Common Mistakes to Avoid

Always check for these common pitfalls, especially under exam pressure:

1. The Sign Error in Substitution:

  • If \((x - 5)\) is a factor, you substitute \(x = +5\).
  • If \((2x - 1)\) is a factor, you substitute \(x = 1/2\). (Remember, set the factor equal to zero to find the value of \(x\)).

2. Missing Terms in Long Division:

If a polynomial is missing a power, such as \(x^3 + 5x - 8\) (no \(x^2\) term), it is highly recommended to write it with a zero placeholder before dividing.
Write it as: \(x^3 + 0x^2 + 5x - 8\). This prevents alignment errors during subtraction.

3. Forgetting the \(x\) root:

If a cubic equation is \(x^3 - 2x^2 = 0\), don't just factor out \(x^2\) and solve. Remember that \(x^2 = 0\) means \(x=0\) is a root twice (two equal roots). Always ensure you find the required number of roots (three for a cubic).

Did you know?

The algebraic long division method you use for polynomials is exactly the same logical process used by computers to divide large binary numbers! It’s a fundamental mathematical procedure.

Quick Review Box

Remainder Theorem:

Dividing \(P(x)\) by \((x - a)\) gives Remainder \(R = P(a)\).

Factor Theorem:

\((x - a)\) is a factor if \(P(a) = 0\).

To Solve \(P(x)=0\) (Cubic):

1. Guess roots \(a\) (factors of constant term).

2. Verify \(P(a)=0\).

3. Divide \(P(x)\) by \((x-a)\) to get the quadratic \(Q(x)\).

4. Solve \(Q(x)=0\) (by factoring or quadratic formula).

You've got this! Practice makes perfect in algebraic manipulation.