The Factor Theorem Revision Notes for Leaving Cert Mathematics

The Factor Theorem

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The Factor Theorem is a fundamental result in algebra that links factors and roots of polynomials. It provides a method to determine whether a linear expression is a factor of a given polynomial.

The number of roots is proportional to the degree of the polynomial. A quadratic equation, with a degree of 2, has two roots. A cubic equation, with a degree of 3 has three roots. This is illustrated from the number of times that equations cross the $x$ -axis.

*quadratic equation *

cubic equation

For a polynomial $P(x)$ :

If $P(c)=0$ for some number $c$ , then $(x-c)$ is a factor of $P(x)$ . The converse is true as well. Example

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The function $f$ is such that $f(x)=2x^3+5x^2-4x-3$ . Show that $x=-3$ is a root of $f(x)$ .

If $x=-3$ is a root, then $f(-3)$ should equal to zero.

f(-3)=2(-3)^3+5(-3)^2-4(-3)-3=0

Therefore, $x=-3$ is a root.

To find the other roots, we could convert the root into a factor.

x=-3 \implies (x+3)

If $(x+3)$ , it means that it divides evenly into $f(x)$ .

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\begin{array}{r|rrrr} x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline \end{array}

First, take the first term in your dividend ( $2x^3$ ) and divide by the first term in your divisor ( $x$ ).

2x^3 / x=2x^2

This is the first term in our quotient, write on top

\begin{array}{r|rrrr} & 2x^2 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline \end{array}

Multiply $2x^2$ with all the terms in the divisor, and write below the line :

\begin{array}{r|rrrr} & 2x^2 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % \end{array}

Switch the sign of these newly multiplied terms.

\begin{array}{r|rrrr} & 2x^2 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & -2x^3 & -6x^2 & & \\ % \end{array}

Add the two columns (the first column always cancels ):

\begin{array}{r|rrrr} & 2x^2 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 \end{array}

The final step (in this cycle) is to bring down the first left term :

\begin{array}{r|rrrr} & 2x^2 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \end{array}

Now the process repeats, take the first term in the most recent line $(-x^2)$ and divide it by the first term in the divisor ( $x$ ). $-x^2 / x=-x$ . This is the next term in the quotient.

\begin{array}{r|rrrr} & 2x^2 & -x \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \end{array}

Multiply this term with all the terms in the divisor :

\begin{array}{r|rrrr} & 2x^2 & -x \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \\ %subtraction & & -x^2 & -3x & \\ % Third \end{array}

Switch signs :

\begin{array}{r|rrrr} & 2x^2 & -x \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \\ %subtraction & & x^2 & 3x & \\ % Third \end{array}

Add lines :

\begin{array}{r|rrrr} & 2x^2 & -x \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \\ %subtraction & & x^2 & 3x & \\ % Third %subtraction & & & -x & \\ % Fourth \end{array}

Bring down the next term :

\begin{array}{r|rrrr} & 2x^2 & -x \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \\ %subtraction & & x^2 & 3x & \\ % Third %subtraction & & & -x & -3 \\ % Fourth \end{array}

Final cycle, take the first term in the most recent line $(-x)$ and divide it by the first term in the divisor ( $x$ ). $-x / x=-1$ . This is the next term in the quotient.

\begin{array}{r|rrrr} & 2x^2 & -x & -1 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \\ %subtraction & & x^2 & 3x & \\ % Third %subtraction & & & -x & -3 \\ % Fourth \end{array}

Multiply with all terms in the divisor :

\begin{array}{r|rrrr} & 2x^2 & -x & -1 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \\ %subtraction & & x^2 & 3x & \\ % Third %subtraction & & & -x & -3 \\ % Fourth %subtraction & & & -x & -3 \\ % Fourth \end{array}

Switch signs :

\begin{array}{r|rrrr} & 2x^2 & -x & -1 \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 &-4x \\ %subtraction & & x^2 & 3x & \\ % Third %subtraction & & & -x & -3 \\ % Fourth %subtraction & & & +x & +3 \\ % Fourth \end{array}

Add lines :

\begin{array}{r|rrrr} & 2x^2 & -x & -1 & \\ % Quotient aligned x + 3 & 2x^3 & +5x^2 & -4x & -3 \\ % Dividend \hline & 2x^3 & +6x^2 & & \\ % First subtraction & & -x^2 & -4x & \\ % Second subtraction & & -x^2 & -3x & \\ % Third subtraction & & & -x & -3 \\ % Fourth subtraction & & & -x & -3 \\ % Final subtraction & & & & 0 \\ % Remainder \end{array}

What is left is a remainder of 0, which makes sense since $(x+3)$ is a factor which should divide evenly into $f(x)$ . What is left is the quotient, which can be used to derive the remaining roots.

2x^2-x-1=0

x=1, x=-\frac{1}{2}

So $f(x)$ can also be expressed as :

(x+3)(x-1)(2x+1)

The Factor Theorem (Leaving Cert Mathematics): Revision Notes