Question

How do you solve the equation $\left( 3x-1 \right)\left( x+4 \right)=0?$

Answer 1

Hint: We will use polynomial multiplication to solve the given equation. We will multiply each term of the first polynomial with every term of the second polynomial. When we multiply the variable terms, we will multiply the coefficients with usual multiplication and the power of the variable changes accordingly. We will use usual multiplication for multiplying the constant terms of the polynomials.

Complete step by step solution:
Let us consider the given polynomial equation $\left( 3x-1 \right)\left( x+4 \right)=0.$
We are asked to solve the given equation.
In order to solve the given equation, we need to find the product of the polynomials on the left-hand side.
We know that $\left( ax+b \right)\left( cx+d \right)=ac{{x}^{2}}+\left( ad+bc \right)x+bd.$
We will use this rule to find the product of the polynomials on the left-hand side of the given polynomial equation.
We will first multiply $3x$ with $x$ to get $3{{x}^{2}}.$
Then we will multiply $3x$ with $4$ to get $12x.$
Let us multiply the constant term of the first polynomial with the terms of the second polynomial.
When $x$ is multiplied with $-1,$ we will get $-x.$
When $4$ is multiplied with $-1,$ we will get $-4.$
So, the product will be $\left( 3x-1 \right)\left( x+4 \right)=3{{x}^{2}}+12x-x-4.$
We will get $\left( 3x-1 \right)\left( x+4 \right)=3{{x}^{2}}+11x-4.$
Let us equate the right-hand side of the equation we have obtained with the right-hand side of the given equation.
We will get $3{{x}^{2}}+11x-4=0.$
This is a quadratic equation. We will use the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for a quadratic equation $a{{x}^{2}}+bx+c=0$ to solve the obtained equation.
When we compare the values, we will get $a=3,b=11$ and $c=4.$
Therefore, the quadratic formula for the given equation is $x=\dfrac{-11\pm \sqrt{{{11}^{2}}-4\times 3\times \left( -4 \right)}}{2\times 3}.$
We will get $x=\dfrac{-11\pm \sqrt{121+48}}{6}=\dfrac{-11\pm \sqrt{169}}{6}=\dfrac{-11\pm 13}{6}.$
We obtain $x=\dfrac{-11-13}{6}=\dfrac{-24}{6}=-4$ or $x=\dfrac{-11+13}{6}=\dfrac{2}{6}=\dfrac{1}{3}.$
Hence the solution is $x=\dfrac{1}{3}$ or $x=-4.$

Note: We can solve the given equation as follows: We are given $\left( 3x-1 \right)\left( x+4 \right)=0.$ So, we know that either $3x-1=0$ or $x+4=0.$ That is, $3x=1$ or $x=-4.$ Therefore, the solution is $x=\dfrac{1}{3}$ or $x=-4.$