2.5.2 Polynomial Division

Definition of a Polynomial:

A polynomial is an expression that contains only positive integer powers of $x$ and constants.
General form: $c_0 + c_1x + c_2x^2 + c_3x^3 + \ldots$

Division Terminology:

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Example: $\frac{7}{2}$

Quotient: $3$
Remainder: $1$
This representation tells us we can "pull out" three whole $2's$ from $7$ , with a single unit being left over at the end. This remaining unit has not yet been divided by $2$ but needs to be.
Therefore, $\frac{7}{2} = 3 + \frac{1}{2}$

Dividing Polynomials:

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Example: $\dfrac{x^2 + 3x + 5}{x + 1}$

Step-by-step process:

Divide the leading term of the dividend $x^2$ by the leading term of the divisor $x$ , which gives $x$ .
Multiply the entire divisor by this result $x \cdot (x + 1) = x^2 + x$ .
Subtract this result from the original polynomial: $(x^2 + 3x + 5) - (x^2 + x) = 2x + 5$ .
Repeat the process with the new polynomial $(2x + 5)$ .
Divide $2x$ by $x$ to get $2$ .
Multiply the divisor by $2$ : $2 \cdot (x + 1) = 2x + 2$ .
Subtract this result from the new polynomial: $(2x + 5) - (2x + 2) = 3$ .

Result:
Quotient: $x + 2$
Remainder: $3$
Therefore, $\dfrac{x^2 + 3x + 5}{x + 1} = x + 2 + \dfrac{3}{x + 1}$

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Example: $\dfrac{x^3 + 5x^2 + 3x-4}{x + 5}$

Step-by-Step Solution:

Setup Polynomial Long Division:

Dividend: $x^3 + 5x^2 + 3x - 4$
Divisor: $x + 5$

First Division:

Divide the leading term of the dividend $(x^3 )$ by the leading term of the divisor $( x )$ , which gives $x^2$ .
Multiply $x^2$ by $x + 5 : x^2 \cdot (x + 5) = x^3 + 5x^2$ .
Subtract $x^3 + 5x^2$ from $x^3 + 5x^2 + 3x - 4$ :
Result: $(x^3 + 5x^2 + 3x - 4) - (x^3 + 5x^2) = 3x - 4$ .

Second Division:

Divide $3x$ by $x$ to get $3$ .
Multiply $3$ by $x + 5 : 3 \cdot (x + 5) = 3x + 15$ .
Subtract $3x + 15$ from $3x - 4$ :
Result: $(3x - 4) - (3x + 15) = -19$ .

Result:

Quotient: $x^2 + 3$
Remainder: $-19$
Therefore, the answer is:
$\dfrac{x^3 + 5x^2 + 3x - 4}{x + 5} = x^2 + 3 - \dfrac{19}{x + 5}$

Final Answer:

$\boxed{x^2 + 3 - \frac{19}{x + 5}}$

Polynomial Problem

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The polynomial $f(x)$ is defined by $f(x) = x^3 - 9x^2 + 7x + 33$ .

(i) Find the remainder when $f(x)$ is divided by $(x + 2)$ .

Using the Remainder Theorem, substitute $x = -2$ into $f(x)$ : $f(-2) = (-2)^3 - 9(-2)^2 + 7(-2) + 33$ $= -8 - 36 - 14 + 33$ $= -25$
Remainder: -25

(ii) Show that $(x - 3)$ is a factor of $f(x)$ .

Using the Factor Theorem, substitute $x = 3$ into $f(x)$ : $f(3) = 3^3 - 9(3)^2 + 7(3) + 33$ $= 27 - 81 + 21 + 33$ $= 0$
Conclusion: x - 3 is a factor of f(x).

(iii) Solve the equation $f(x) = 0$ , giving each root in exact form as simply as possible.

Perform polynomial division of $f(x)$ by $x - 3$ : $\begin{array}{r|rrrr} & x^3 & - 9x^2 & + 7x & + 33 \\ x-3 & x^2 & - 6x & - 11 \\ \end{array}$
Step-by-step division:
- Divide $x^3$ by $x$ : $x^2$
- Multiply $x^2$ by $x - 3: x^3 - 3x^2$
- Subtract: $(x^3 - 9x^2 + 7x + 33) - (x^3 - 3x^2) = -6x^2 + 7x + 33$
- Divide $-6x^2$ by $x: -6x$
- Multiply $-6x$ by $x - 3: -6x^2 + 18x$
- Subtract: $(-6x^2 + 7x + 33) - (-6x^2 + 18x) = -11x + 33$
- Divide $-11x$ by $x: -11$
- Multiply $-11$ by $x - 3 : -11x + 33$
- Subtract: $(-11x + 33) - (-11x + 33) = 0$
Quotient: $x^2 - 6x - 11$
Therefore, $f(x) = (x - 3)(x^2 - 6x - 11)$
Solve $x^2 - 6x - 11 = 0$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 1, \quad b = -6, \quad c = -11$ $x = \frac{6 \pm \sqrt{36 + 44}}{2}$ $x = \frac{6 \pm \sqrt{80}}{2}$ $x = \frac{6 \pm 4\sqrt{5}}{2}$ $x = 3 \pm 2\sqrt{5}$
$\boxed {\therefore \quad x = 3, \quad x = 3 \pm 2\sqrt{5}}$

Further Polynomial Division

In Year 1, polynomial division only occurred with a linear denominator.

In Year 2, the order of the denominator could be linear or greater.

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Example: Simplify $\dfrac{x^4 + 3x^3 + 9x - 12}{x+2}$

Set up the division:

\frac{x^4 + 3x^3 + 9x - 12}{x + 2}

\begin{array}{c|c} & x^3 &-2x^2 & +7x & -5 \\ \hline X & x^4 & -2x^3 & +7x^2 & -5x \\ +2 & 2x^3 & -4x^2 & +14x & -10 & \quad -2\\ \end{array}

Start dividing:

Divide the first term of the numerator by the first term of the denominator: $\frac{x^4}{x} = x^3$ .
Multiply $x^3 by (x + 2)$ and subtract from the numerator:

x^4 - x^4 + 2x^3 = 3x^3 - 2x^3 = x^3

Divide $x^3$ by $x: \frac{x^3}{x} = x^2$ .
Multiply $x^2$ by $(x + 2)$ and subtract:

x^3 - x^3 + 2x^2 = 3x^2 - 2x^2 = x^2

Continue this process:

x^3 - 2x^2 + 7x - 5 - \text{(quotient and remainder calculations)}

The quotient ( $Q$ ) and remainder ( $r$ ) are:

Q = x^3 - 2x^2 + 7x - 5

r = -2

Write the final answer:

x^3 - 2x^2 + 7x - 5 + \frac{-2}{x+2}

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Division Example: $\dfrac {2x^4-29x^2-40x-40}{x^2-3x-8}$

\begin{array}{c|c} & 2x^2&+6x & +5 \\ \hline X^2 & 2x^4 & +6x^3&5x^2 \\ -3x & -6x^2 & -18x^2 &-15x & \quad +23x\\ -8&-16x^2 &-48x &-40 \end{array}

Must add Up $-40x$ and $-15x$

to $-40x \therefore$ $r=23x$

Need $-40$ in answer $\therefore$ no constant reminder

2x^2+6x+5+\dfrac {23x}{x^2-3x-8}

Remainder and Conclusion:

The remainder after the division is $23x$ , and there is no constant remainder because the constant $−40$ is already accounted for in the original polynomial.
Therefore, the quotient is $2x^2+6x+5$ with a remainder of $\frac{23x}{x^2 - 3x - 8}$ .

Polynomial Division Simplified Revision Notes for A-Level OCR Maths Pure

2.5.2 Polynomial Division

Definition of a Polynomial:

Division Terminology:

Dividing Polynomials:

Example: $\dfrac{x^3 + 5x^2 + 3x-4}{x + 5}$

Step-by-Step Solution:

Final Answer:

Polynomial Problem

(i) Find the remainder when $f(x)$ is divided by $(x + 2)$ .

(ii) Show that $(x - 3)$ is a factor of $f(x)$ .

(iii) Solve the equation $f(x) = 0$ , giving each root in exact form as simply as possible.

Further Polynomial Division

Example: Simplify $\dfrac{x^4 + 3x^3 + 9x - 12}{x+2}$

Remainder and Conclusion:

500K+ Students Use These Powerful Tools to Master Polynomial Division For their A-Level Exams.

Other Revision Notes related to Polynomial Division you should explore

Polynomial Division Simplified Revision Notes for A-Level OCR Maths Pure

2.5.2 Polynomial Division

Definition of a Polynomial:

Division Terminology:

Dividing Polynomials:

Example: x3+5x2+3x−4x+5\dfrac{x^3 + 5x^2 + 3x-4}{x + 5}x+5x3+5x2+3x−4​

Step-by-Step Solution:

Final Answer:

Polynomial Problem

(i) Find the remainder when f(x)f(x)f(x) is divided by (x+2)(x + 2)(x+2).

(ii) Show that (x−3)(x - 3)(x−3) is a factor of f(x)f(x)f(x).

(iii) Solve the equation f(x)=0f(x) = 0f(x)=0, giving each root in exact form as simply as possible.

Further Polynomial Division

Example: Simplify x4+3x3+9x−12x+2\dfrac{x^4 + 3x^3 + 9x - 12}{x+2}x+2x4+3x3+9x−12​

Remainder and Conclusion:

500K+ Students Use These Powerful Tools to Master Polynomial Division For their A-Level Exams.

Other Revision Notes related to Polynomial Division you should explore

Example: $\dfrac{x^3 + 5x^2 + 3x-4}{x + 5}$

(i) Find the remainder when $f(x)$ is divided by $(x + 2)$ .

(ii) Show that $(x - 3)$ is a factor of $f(x)$ .

(iii) Solve the equation $f(x) = 0$ , giving each root in exact form as simply as possible.

Example: Simplify $\dfrac{x^4 + 3x^3 + 9x - 12}{x+2}$