Factor Theorem (AQA A-Level Mathematics): Revision Notes

Example:

• Find the remainder when $x^3 + 8x^2 - 6x + 2$ is divided by $x + 6$.

1. Polynomial Long Division:

• Divide $x^3 + 8x^2 - 6x + 2$ by $x + 6$.

• Step-by-step division process yields a quotient and a remainder.

• Result:

• Quotient: $x^2 + 2x - 18$

• Remainder: 110

• Therefore, $r = 110$

2. Using the Remainder Theorem:

• A quicker way to find the remainder is using the remainder theorem.
• If $f(x)$ is divided by $x - c$, the remainder is $f(c)$.
• Here, $f(x) = x^3 + 8x^2 - 6x + 2$, and we divide by $x + 6 ( x - (-6) )$.
• Substituting $x = -6$ into $f(x)$:
• $f(-6) = (-6)^3 + 8(-6)^2 - 6(-6) + 2$
• $f(-6) = -216 + 288 + 36 + 2$
• $f(-6) = 110$

3. Conclusion:

• The remainder when $x^3 + 8x^2 - 6x + 2$ is divided by $x + 6$ is 110.

Example Problem:

Given: $g(x) = 2x^3 + x^2 - 13x + 6$

Task:

4. Show that $(x + 3)$ is a factor of $g(x)$.
5. Solve the equation $g(x) = 0$.

Solution:

6. Check if $(x + 3)$ is a factor using the Factor Theorem:

• Evaluate $g(-3)$: $g(-3) = 2(-3)^3 + (-3)^2 - 13(-3) + 6$ $g(-3) = 2(-27) + 9 + 39 + 6$ $g(-3) = -54 + 9 + 39 + 6$ $g(-3) = 0$
• Since $g(-3) = 0$, $x + 3$ is a factor by the Factor Theorem.

7. Find the other factors:

• Perform polynomial division of $g(x)$ by $x + 3$ : $\begin{array}{r|rrrr} & 2x^3 & + x^2 & - 13x & + 6 \\ x+3 & 2x^2 & -5x & + 2 & \\ \end{array}$
• Step-by-step division:
• Divide $2x^3$ by $x : 2x^2$
• Multiply $2x^2$ by $x + 3 : 2x^3 + 6x^2$
• Subtract: $2x^3 + x^2 - 13x + 6) - (2x^3 + 6x^2) = -5x^2 - 13x + 6$
• Divide -$5x^2$ by $x : -5x$
• Multiply $-5x$ by $x + 3: -5x^2 - 15x$
• Subtract: $(-5x^2 - 13x + 6) - (-5x^2 - 15x) = 2x + 6$
• Divide $2x$ by $x : 2$
• Multiply $2$ by $x + 3 : 2x + 6$
• Subtract: $(2x + 6) - (2x + 6) = 0$
• Therefore: $g(x) = (x + 3)(2x^2 - 5x + 2)$

8. Factor the quadratic $2x^2 - 5x + 2$:

• Use the quadratic formula $ax^2 + bx + c$: $2x^2 - 5x + 2 = 0$
• Solution using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2}$ $x = \frac{5 \pm \sqrt{25 - 16}}{4}$ $x = \frac{5 \pm 3}{4}$ $x = 2 \quad \text{or} \quad x = \frac{1}{2}$
• Therefore: $2x^2 - 5x + 2 = (2x - 1)(x - 2)$

9. Complete Factorization and Solution: $g(x) = (x + 3)(2x - 1)(x - 2)$

• Solving $g(x) = 0$: $(x + 3)(2x - 1)(x - 2) = 0$ $x = -3, \quad 2x =1\quad x = \frac{1}{2}, \quad x = 2$
• Therefore, the solutions are: $\boxed {x = -3, \frac{1}{2}, 2}$

Example 1: Finding the Remainder

• Problem: Find the remainder when $x^3 + 6x^2 + 2$ is divided by $3x - 2$.
• Set $3x - 2 = 0$ to find $x = \frac{2}{3}$.
• Adapt the remainder theorem: Substitute $x = \frac{2}{3}$ into the polynomial.
• Calculation:

$f\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 + 6\left(\frac{2}{3}\right)^2 + 2$ $= \frac{8}{27} + 6 \cdot \frac{4}{9} + 2$ $= \frac{8}{27} + \frac{24}{9} + 2$ $= \frac{8}{27} + \frac{72}{27} + \frac{54}{27}$ $= \frac{134}{27}$

• Remainder: $\boxed {r = \frac{134}{27}}$

Example 2: Factorization

• Problem: Show that $(x - 2)$ is a factor of $2x^3 + 3x^2 - 18x + 8$ and factorise completely.
• Use the factor theorem: Substitute $x = 2$ and check if $f(2) = 0$.
• Calculation:

$f(2) = 2(2)^3 + 3(2)^2 - 18(2) + 8$ $= 2(8) + 3(4) - 36 + 8$ $= 16 + 12 - 36 + 8$ $= 0$

• Conclusion: $x - 2$ is a factor.
• Finding Other Factors:
• Polynomial division: $\begin{array}{r|rrrr} & 2x^3 & + 3x^2 & - 18x & + 8 \\ x-2 & 2x^2 & + 7x & - 4 & \\ \end{array}$
• Dividing: $2x^3 - 18x + 8 = (x-2)(2x^2 + 7x - 4)$
• Further factorise $2x^2 + 7x - 4$: $2x^2 + 7x - 4 = (2x - 1)(x + 4)$
• Complete Factorization: $\boxed {g(x) = (x - 2)(2x - 1)(x + 4)}$

Example 3: Finding Values of $c$ and $d$

• Problem: Given that $(x + 1)$ and $(x - 2)$ are factors of $cx^3 + dx^2 - 9x - 10$, find $c$ and $d$.

• Conditions: $f(-1) = 0 \quad \text{and} \quad f(2) = 0$

• Calculation:

• For $f(-1) = 0$ : $c(-1)^3 + d(-1)^2 - 9(-1) - 10 = 0$ $=-c + d + 9 - 10 = 0$ $-c + d - 1 = 0 \quad \Rightarrow \quad -c + d = 1$

• For $f(2) = 0$: $c(2)^3 + d(2)^2 - 9(2) - 10 = 0$ $=8c + 4d - 18 - 10 = 0$ $8c + 4d - 28 = 0 \quad \\ \Rightarrow \quad 2c + d = 7$

• Solving the System:

  

• Use simultaneous equations: $\begin{cases} -c + d = 1 \\ 2c + d = 7 \end{cases}$ Subtract the first equation from the second: $3c = 6 \quad \Rightarrow \quad c = 2$

• Substitute $c = 2$ into the first equation: $-2 + d = 1 \quad \Rightarrow \quad d = 3$

• Solution: $\boxed {c = 2,\\ \quad d = 3}$