Question

How do you solve $\left( 2m+3 \right)\left( 4m+3 \right)=0$ ? \[\]

Answer 1

Hint: We recall factors of a polynomial. The solutions of the given equation will be zeros of the polynomial at the left hand side. We recall if the product of two factors is zero then at least one of them is zero. We equate the factors $\left( 2m+3 \right),\left( 4m+3 \right)$to zero separately and solve for $m$. \[\]

Complete step by step answer:
We know that when we divide a divided polynomial $p\left( x \right)$ with degree $n$ by some divisor polynomial $d\left( x \right)$ with degree $m\le n$ then we get the quotient polynomial $q\left( x \right)$ of degree $n-m$ and the remainder polynomial as $r\left( x \right)$.We use Euclidean division formula and can write as
\[ p\left( x \right)=d\left( x \right)q\left( x \right)+r\left( x \right)\]
We also know that if the remainder polynomial is zero that is $r\left( x \right)=0$ then we call $d\left( x \right),q\left( x \right)$ factor polynomial of $p\left( x \right)$ or simply factors of $p\left( x \right)$. We are given the following equation with unknown variable $m$as
\[\left( 2m+3 \right)\left( 4m+3 \right)=0\]
We see that in the left hand side there are two factors $2m+3,4m+3$ are multiplied to each other and their product is at right hand side zero. We know that if product of two factors is zero then at least one of them is zero. So we have $2m+3=0\text{ or }4m+3=0$. Let us solve one by one.
\[2m+3=0\]
We subtract both sides of the above equation by 3 to have;
\[\begin{align}
  & 2m+3-3=0-3 \\
 & \Rightarrow 2m=-3 \\
\end{align}\]
We divide both sides of above equation by 2 to have
\[\Rightarrow m=\dfrac{-3}{2}....\left( 1 \right)\]
\[4m+3=0\]
We subtract both sides of the above equation by 3 to have;
\[\begin{align}
  & 4m+3-3=0-3 \\
 & \Rightarrow 4m=-3 \\
\end{align}\]
We divide both sides of above equation by 4 to have
\[\Rightarrow m=\dfrac{-3}{4}....\left( 2 \right)\]
We see from (1) and (2) the solutions to the given equation are
\[\Rightarrow m=\dfrac{-3}{2},\dfrac{-3}{4}\]

Note: We alternatively solve by first expanding the left hand side as
\[\begin{align}
  & \Rightarrow \left( 2m+3 \right)\left( 4m+3 \right)=0 \\
 & \Rightarrow 2m\cdot 4m+2m\cdot 3+3\cdot 4m+3\cdot 3=0 \\
 & \Rightarrow 8{{m}^{2}}+18m+9=0 \\
\end{align}\]
We see that the above equation is a quadratic equation in $m$. We solve using the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to have;
\[\begin{align}
  & \Rightarrow m=\dfrac{-18\pm \sqrt{{{18}^{2}}-4\cdot 8\cdot 9}}{2\cdot 8} \\
 & \Rightarrow m=\dfrac{-18\pm \sqrt{324-288}}{16} \\
 & \Rightarrow m=\dfrac{-18\pm \sqrt{36}}{16} \\
 & \Rightarrow m=\dfrac{-18+6}{16},\dfrac{-18-6}{16} \\
 & \Rightarrow m=\dfrac{-3}{4},\dfrac{-3}{2} \\
\end{align}\]