Question

Solve the equation $8{{x}^{3}}-20{{x}^{2}}+6x+9=0$ given that the equation has multiple roots.

Answer 1

We need to solve the equation $8{{x}^{3}}-20{{x}^{2}}+6x+9=0$. Since the equation is of degree $3$, we need to find the factor of this equation using the trial and error method. The value of $x$ for which the given polynomial is zero will form one root and hence a factor will be obtained. Then divide the given polynomial by this factor using a long division method to get a quotient and remainder and write in the form $Dividend=divisor\times quotient+remainder$. Now rearrange the terms to get two more other factors. By substituting these factors to $0$, the values of $x$ will be obtained.

Complete step-by-step solution
We need to solve the equation $8{{x}^{3}}-20{{x}^{2}}+6x+9=0$.
Let $p(x)=8{{x}^{3}}-20{{x}^{2}}+6x+9$ .
First, we have to find the factor of this equation. Since the given equation is of degree 3, there will be 3 factors and hence 3 roots.
The first root can be found out using the trial and error method, i.e. substitute some value for $x$ such that the result will be a zero.
Let us check the value of the equation at $x=-\dfrac{1}{2}$ . So the given equation will be written as
$p\left( -\dfrac{1}{2} \right)=8\times {{\left( -\dfrac{1}{2} \right)}^{3}}-20\times {{\left( -\dfrac{1}{2} \right)}^{2}}+6\times \left( -\dfrac{1}{2} \right)+9$
By simplifying the above equation, we get
$p\left( -\dfrac{1}{2} \right)=8\times -\dfrac{1}{8}-20\times \dfrac{1}{4}-3+9$
When we solve this, we get
$p\left( -\dfrac{1}{2} \right)=-1-5+6=-1+1$
$\Rightarrow p\left( -\dfrac{1}{2} \right)=0$
Therefore, one factor of the given polynomial is $\left( x+\dfrac{1}{2} \right)$ .
Now, let us divide $p(x)=8{{x}^{3}}-20{{x}^{2}}+6x+9$by $\left( x+\dfrac{1}{2} \right)$ .
Step 1. Let us divide $8{{x}^{3}}$by $\left( x+\dfrac{1}{2} \right)$ .
We will get $8{{x}^{2}}$ as quotient . Now multiply this with $\left( x+\dfrac{1}{2} \right)$ to get $8{{x}^{3}}+4{{x}^{2}}$.
Let us subtract $8{{x}^{3}}+4{{x}^{2}}$ from $8{{x}^{3}}-20{{x}^{2}}+6x+9$ ,i.e
$8{{x}^{3}}-20{{x}^{2}}+6x+9-8{{x}^{3}}-4{{x}^{2}}=-24{{x}^{2}}+6x+9$
Step2: We should divide $-24{{x}^{2}}+6x+9$ by $\left( x+\dfrac{1}{2} \right)$ . The quotient will be $-24x$ and multiplying this with $\left( x+\dfrac{1}{2} \right)$ we will get $-24{{x}^{2}}-12x$ .
Now, we can subtract $-24{{x}^{2}}-12x$ from $-24{{x}^{2}}+6x+9$ , i.e
$-24{{x}^{2}}+6x+9+24{{x}^{2}}+12x=18x+9$
Step3: Let us divide $18x+9$ by $\left( x+\dfrac{1}{2} \right)$ . The quotient will be $18$ and multiplying this with $\left( x+\dfrac{1}{2} \right)$ we will get $18x+9$ .
Now, let subtract $18x+9$ from $18x+9$ , i.e
$18x+9-18x-9=0$
That is, the remainder is $0$ .

8{x^3} - 20{x^2} + 6x + 9\\\ 8{x^3} - 4{x^2}\\\ \begin{array}{*{20}{c}} \- & \+ \end{array}\\\ \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\\ \begin{array}{*{20}{c}} {}&\begin{array}{l} \- 24{x^2} + 6x + 9\\\ \- 24{x^2} - 12x\\\ \begin{array}{*{20}{c}} \+ &{}& \+ \end{array}\\\ \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\\ \begin{array}{*{20}{c}} {}&{}&\begin{array}{l} 18x + 9\\\ 18x + 9\\\ \begin{array}{*{20}{c}} \- & \- \end{array}\\\ \\_\\_\\_\\_\\_\\_\\_\\_\\_\\\ \begin{array}{*{20}{c}} {}&{}&0 \end{array} \end{array} \end{array} \end{array} \end{array} \end{array}}}\right. \\!\\!\\!\\!\overline{\,\,\,\vphantom 1{\begin{array}{l} 8{x^3} - 20{x^2} + 6x + 9\\\ 8{x^3} - 4{x^2}\\\ \begin{array}{*{20}{c}} \- & \+ \end{array}\\\ \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\\ \begin{array}{*{20}{c}} {}&\begin{array}{l} \- 24{x^2} + 6x + 9\\\ \- 24{x^2} - 12x\\\ \begin{array}{*{20}{c}} \+ &{}& \+ \end{array}\\\ \\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\_\\\ \begin{array}{*{20}{c}} {}&{}&\begin{array}{l} 18x + 9\\\ 18x + 9\\\ \begin{array}{*{20}{c}} \- & \- \end{array}\\\ \\_\\_\\_\\_\\_\\_\\_\\_\\_\\\ \begin{array}{*{20}{c}} {}&{}&0 \end{array} \end{array} \end{array} \end{array} \end{array} \end{array}}}} \limits^{\displaystyle\,\,\, {8{x^2} - 24x + 18}}$$ So, when we divide $p(x)=8{{x}^{3}}-20{{x}^{2}}+6x+9$ by $\left( x+\dfrac{1}{2} \right)$ we get the quotient as $8{{x}^{2}}-24x+18$ . Now, $Dividend=divisor\times quotient+remainder$ That is, $p(x)=\left( x+\dfrac{1}{2} \right)\times \left( 8{{x}^{2}}-24x+18 \right)+0$ $p(x)=\left( x+\dfrac{1}{2} \right)\times \left( 8{{x}^{2}}-24x+18 \right)$ Let us take 2 outside, we will get $=\left( x+\dfrac{1}{2} \right)\times 2\left( 4{{x}^{2}}-12x+9 \right)$ $=\left( x+\dfrac{1}{2} \right)\times 2\left( {{\left( 2x \right)}^{2}}-2\times 2\times 3x+{{3}^{2}} \right)$ ${{\left( 2x \right)}^{2}}-2\times 2\times 3x+{{3}^{2}}$ is of the form ${{(a+b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . Therefore, we can write the above equation as $p(x)=\left( x+\dfrac{1}{2} \right)\times 2{{\left( 2x+3 \right)}^{2}}$ $=\left( \dfrac{2x+1}{2} \right)\times 2{{\left( 2x+3 \right)}^{2}}$ Let us cancel $2$ from numerator and denominator, we get $p(x)=\left( 2x+1 \right){{\left( 2x+3 \right)}^{2}}$ i.e. $p(x)=\left( 2x+1 \right)\left( 2x+3 \right)\left( 2x+3 \right)$ Now, we should substitute $p(x)=0$ . i.e. $\left( 2x+1 \right)\left( 2x+3 \right)\left( 2x+3 \right)=0$ $\Rightarrow \left( 2x+1 \right)=0,\left( 2x+3 \right)=0,\left( 2x+3 \right)=0$ Let us evaluate the value of $x$ . $\left( 2x+1 \right)=0\Rightarrow x=\dfrac{-1}{2}$ $\left( 2x+3 \right)=0\Rightarrow x=\dfrac{-3}{2}$ **Hence the value of $x$ are $\dfrac{-1}{2},\dfrac{-3}{2},\dfrac{-3}{2}$.** **Note:** In questions having degree 3 or more, the same procedure is followed starting from the trial and error method to obtain a factor and doing the long division method and simplifying. Note that the result obtained by doing the trial and error method must be 0 and the value of $x$ chosen for the same, say $a$ must be written in the form $x+a$ if $a$ is negative and $x-a$ if $a$ is positive.