3.1 Factor theorem exercises, solving cubic equations

Siyavula textbooks: grade 12 Page 1 / 1

Exercises - using factor theorem

Find the remainder when $4 x^{3} - 4 x^{2} + x - 5$ is divided by $(x + 1)$ .
Use the factor theorem to factorise $x^{3} - 3 x^{2} + 4$ completely.
f ( x ) = 2 x 3 + x 2 - 5 x + 2
1. Find $f (1)$ .
2. Factorise $f (x)$ completely
Use the Factor Theorem to determine all the factors of the following expression:
$x^{3} + x^{2} - 17 x + 15$
Complete: If $f (x)$ is a polynomial and $p$ is a number such that $f (p) = 0$ , then $(x - p)$ is .....

Solving cubic equations

Once you know how to factorise cubic polynomials, it is also easy to solve cubic equations of the kind

a x^{3} + b x^{2} + c x + d = 0

Solve

6 x^{3} - 5 x^{2} - 17 x + 6 = 0 .

Find one factor using the Factor Theorem
Try

$f (1) = 6 {(1)}^{3} - 5 {(1)}^{2} - 17 (1) + 6 = 6 - 5 - 17 + 6 = - 10$

Therefore $(x - 1)$ is NOT a factor.

Try

$f (2) = 6 {(2)}^{3} - 5 {(2)}^{2} - 17 (2) + 6 = 48 - 20 - 34 + 6 = 0$

Therefore $(x - 2)$ IS a factor.
Division by inspection
$6 x^{3} - 5 x^{2} - 17 x + 6 = (x - 2) ()$

The first term in the second bracket must be $6 x^{2}$ to give $6 x^{3}$ if one works backwards.

The last term in the second bracket must be $- 3$ because $- 2 \times - 3 = + 6$ .

So we have $6 x^{3} - 5 x^{2} - 17 x + 6 = (x - 2) (6 x^{2} + ? x - 3)$ .

Now, we must find the coefficient of the middle term ( $x$ ).

$(- 2) (6 x^{2})$ gives $- 12 x^{2}$ . So, the coefficient of the $x$ -term must be 7.

So, $6 x^{3} - 5 x^{2} - 17 x + 6 = (x - 2) (6 x^{2} + 7 x - 3)$ .
Factorise fully
$6 x^{2} + 7 x - 3$ can be further factorised to $(2 x + 3) (3 x - 1)$ ,

and we are now left with $6 x^{3} - 5 x^{2} - 17 x + 6 = (x - 2) (2 x + 3) (3 x - 1)$
Solve the equation
$\begin{matrix} 6 x^{3} - 5 x^{2} - 17 x + 6 & = & 0 \\ (x - 2) (2 x + 3) (3 x - 1) & = & 0 \\ x & = & 2; \frac{1}{3}; - \frac{3}{2} \end{matrix}$

Sometimes it is not possible to factorise the trinomial ("second bracket"). This is when the quadratic formula

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

can be used to solve the cubic equation fully.

For example:

Solve for $x$ : $x^{3} - 2 x^{2} - 6 x + 4 = 0 .$

Find one factor using the Factor Theorem
Try

$f (1) = {(1)}^{3} - 2 {(1)}^{2} - 6 (1) + 4 = 1 - 2 - 6 + 4 = - 1$

Therefore $(x - 1)$ is NOT a factor.

Try

$f (2) = {(2)}^{3} - 2 {(2)}^{2} - 6 (2) + 4 = 8 - 8 - 12 + 4 = - 8$

Therefore $(x - 2)$ is NOT a factor.

$f (- 2) = {(- 2)}^{3} - 2 {(- 2)}^{2} - 6 (- 2) + 4 = - 8 - 8 + 12 + 4 = 0$

Therefore $(x + 2)$ IS a factor.
Division by inspection
$x^{3} - 2 x^{2} - 6 x + 4 = (x + 2) ()$

The first term in the second bracket must be $x^{2}$ to give $x^{3}$ .

The last term in the second bracket must be 2 because $2 \times 2 = + 4$ .

So we have $x^{3} - 2 x^{2} - 6 x + 4 = (x + 2) (x^{2} + ? x + 2)$ .

Now, we must find the coefficient of the middle term ( $x$ ).

$(2) (x^{2})$ gives $2 x^{2}$ . So, the coefficient of the $x$ -term must be $- 4$ . ( $2 x^{2} - 4 x^{2} = - 2 x^{2}$ )

So $x^{3} - 2 x^{2} - 6 x + 4 = (x + 2) (x^{2} - 4 x + 2)$ .

$x^{2} - 4 x + 2$ cannot be factorised any futher and we are now left with

$(x + 2) (x^{2} - 4 x + 2) = 0$
Solve the equation
$\begin{matrix} (x + 2) (x^{2} - 4 x + 2) & = & 0 \\ (x + 2) = 0 & o r & (x^{2} - 4 x + 2) = 0 \end{matrix}$
Apply the quadratic formula for the second bracket
Always write down the formula first and then substitute the values of $a, b$ and $c$ .

$\begin{matrix} x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = & \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (2)}}{2 (1)} \\ = & \frac{4 \pm \sqrt{8}}{2} \\ = & 2 \pm \sqrt{2} \end{matrix}$
Final solutions
$x = - 2 o r x = 2 \pm \sqrt{2}$

Exercises - solving of cubic equations

Solve for $x$ : $x^{3} + x^{2} - 5 x + 3 = 0$
Solve for $y$ : $y^{3} - 3 y^{2} - 16 y - 12 = 0$
Solve for $m$ : $m^{3} - m^{2} - 4 m - 4 = 0$
Solve for $x$ : $x^{3} - x^{2} = 3 (3 x + 2)$
:
Remove brackets and write as an equation equal to zero.
Solve for $x$ if $2 x^{3} - 3 x^{2} - 8 x = 3$

End of chapter exercises

Solve for $x$ : $16 (x + 1) = x^{2} (x + 1)$
1. Show that $x - 2$ is a factor of $3 x^{3} - 11 x^{2} + 12 x - 4$
2. Hence, by factorising completely, solve the equation
  $3 x^{3} - 11 x^{2} + 12 x - 4 = 0$
$2 x^{3} - x^{2} - 2 x + 2 = Q (x) . (2 x - 1) + R$ for all values of $x$ . What is the value of $R$ ?
1. Use the factor theorem to solve the following equation for $m$ :
  $8 m^{3} + 7 m^{2} - 17 m + 2 = 0$
2. Hence, or otherwise, solve for $x$ :
  $2^{3 x + 3} + 7 \cdot 2^{2 x} + 2 = 17 \cdot 2^{x}$
A challenge : Determine the values of $p$ for which the function
$f (x) = 3 p^{3} - (3 p - 7) x^{2} + 5 x - 3$
leaves a remainder of 9 when it is divided by $(x - p)$ .

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Source: OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2

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