Chapter SH, Problem 8.4EP

To determine

To simplify: the given expression using proper mathematical formulae.

Answer to Problem 8.4EP

  $\frac{{\text{x}}^{\text{2}}\text{-2x-15}}{{\text{x}}^{\text{2}}\text{-6x+8}}$ .

Explanation of Solution

Given:

  $\frac{{\text{x}}^{\text{2}}\text{+10x+21}}{{\text{x}}^{\text{2}}\text{-5x+6}}\text{÷}\frac{{\text{x}}^{\text{2}}\text{+3x-28}}{{\text{x}}^{\text{2}}\text{-8x+15}}$ .

Concept used:

The second term will reciprocal if division changed to multiplication.

Factorization of quadratic polynomials of the form ${\text{x}}^{\text{2}}\text{+bx+c}$ .

In order to factorize ${\text{x}}^{\text{2}}\text{+bx+c}$ first find the two numbers $\text{p}$ and $\text{q}$ such that $\text{p+q=b}$ and $\text{pq=c}$ .

After finding $\text{p}$ and $\text{q}$ , split the middle term in the quadratic as $\text{px+qx}$ and get desired factors by grouping the terms.

Calculation:

  $\frac{{\text{x}}^{\text{2}}\text{+10x+21}}{{\text{x}}^{\text{2}}\text{-5x+6}}\text{÷}\frac{{\text{x}}^{\text{2}}\text{+3x-28}}{{\text{x}}^{\text{2}}\text{-8x+15}}$ .

  $\Rightarrow \frac{{\text{x}}^{\text{2}}\text{+10x+21}}{{\text{x}}^{\text{2}}\text{-5x+6}}\times \frac{{\text{x}}^{\text{2}}\text{-8x+15}}{{\text{x}}^{\text{2}}\text{+3x-28}}$ .

Taking the middle term factorization:

  $\frac{{\text{x}}^{\text{2}}\text{+7x+3x+21}}{{\text{x}}^{\text{2}}\text{-3x-2x+6}}\times \frac{{\text{x}}^{\text{2}}\text{-3x-5x+15}}{{\text{x}}^{\text{2}}\text{+7x-4x-28}}$ .

  $\Rightarrow \frac{\text{x}\left(\text{x+7}\right)\text{+3}\left(\text{x+7}\right)}{\text{x}\left(\text{x-3}\right)\text{-2}\left(\text{x-3}\right)}\text{×}\frac{\text{x}\left(\text{x-3}\right)\text{-5}\left(\text{x-3}\right)}{\text{x}\left(\text{x+7}\right)\text{-4}\left(\text{x+7}\right)}$ .

  $\Rightarrow \frac{\left(\text{x+7}\right)\left(\text{x+3}\right)}{\left(\text{x-2}\right)\left(\text{x-3}\right)}\text{×}\frac{\left(\text{x-5}\right)\left(\text{x-3}\right)}{\left(\text{x-4}\right)\left(\text{x+7}\right)}$ .

Dividing numerator and denominator both by $\left(\text{x+7}\right)\left(\text{x-3}\right)$ :

  $\frac{\left(\text{x+3}\right)\left(\text{x-5}\right)}{\left(\text{x-2}\right)\left(\text{x-4}\right)}$ .

  $\Rightarrow \frac{{\text{x}}^{\text{2}}\text{-5x+3x-15}}{{\text{x}}^{\text{2}}\text{-4x-2x+8}}$ .

Hence, the expression can be simplified to $\frac{{\text{x}}^{\text{2}}\text{-2x-15}}{{\text{x}}^{\text{2}}\text{-6x+8}}$ .