Assume that the width of the rectangle = x
Since the length is twice the sum of the width and 3, then
[tex]\begin{gathered} L=2(x+3) \\ L=2x+6 \end{gathered}[/tex]
Since the area of the rectangle is 216 square inches, then
Multiply the length and the width, then equate the product by 216
[tex]\begin{gathered} (x)(2x+6)=216 \\ 2x^2+6x=216 \end{gathered}[/tex]
Divide all terms by 2 to simplify
[tex]\begin{gathered} \frac{2x^2}{2}+\frac{6x}{2}=\frac{216}{2} \\ x^2+3x=108 \end{gathered}[/tex]
Subtract 108 from both sides
[tex]\begin{gathered} x^2+3x-108=108-108 \\ x^2+3x-108=0 \end{gathered}[/tex]
Now, let us factorize the trinomial into 2 factors
[tex]\begin{gathered} x^2=(x)(x) \\ -108=(-9)(12) \\ (x)(-9)+(x)(12)=-9x+12x=3x \end{gathered}[/tex]
Then the factors are
[tex]x^2+3x-108=^{}(x-9)(x+12)[/tex]
Equate them by 0
[tex](x-9)(x+12)=0[/tex]
Equate each factor by 0, then find the values of x
[tex]x-9=0[/tex]
Add 9 to both sides
[tex]\begin{gathered} x-9+9=0+9 \\ x=9 \end{gathered}[/tex][tex]x+12=0[/tex]
Subtract 12 from both sides
[tex]\begin{gathered} x+12-12=0-12 \\ x=-12 \end{gathered}[/tex]
Since the width can not be a negative number (no negative length)
Then the width of the rectangle = 9
Let us find the length
[tex]\begin{gathered} L=2(9)+6 \\ L=18+6 \\ L=24 \end{gathered}[/tex]
Then the dimensions of the rectangle are 9 inches and 24 inches
