Chapter SH, Problem 14.8.11EP

(a)

To determine

To find: the dimensions of the given figure.

Answer to Problem 14.8.11EP

Hence, the result is $\left(x-4\right)\left(x+3\right)$

Explanation of Solution

Given: $\left(x^2-x-12\right)$ .

  

Calculation:

Factor the below trinomial $x^2-x-12$

Write the trinomial as the product of two binomials.

Product of first terms $x^2$ .

  $x^2-x-12=\left(x+?\right)\left(x+?\right)$

Now find the two last terms whose product must be $-12$ .

Try several factors of $-12$ so that sum of the product of outer and inner terms must be $-x$ .

Try the factors $-4,+3$ .

  $\begin{array}{l}\left(x-4\right)\left(x+3\right)=x^2-4x+3x-12\\ =x^2-x-12\end{array}$

Hence, the result is $\left(x-4\right)\left(x+3\right)$

Conclusion:

Hence, the result is $\left(x-4\right)\left(x+3\right)$

(b)

To determine

To find: the area of the figure

Answer to Problem 14.8.11EP

Hence, the required area is $60m^2$ .

Explanation of Solution

Calculation:

Find the area if $x=9$ .

Substitute $x=9$ in the above result.

  $\begin{array}{l}\left(x-4\right)\left(x+3\right)=\left(9-4\right)\left(9+3\right)\\ =5\times 12\\ =60\end{array}$

Hence, the required area is $60m^2$ .

Conclusion:

Hence, the required area is $60m^2$ .