Chapter SH, Problem 8.5EP

To determine

To simplify: the function of average hourly income of college student after the graduation.

Answer to Problem 8.5EP

  $\text{I}\left(\text{x}\right)\text{=11x+11}$ .

Explanation of Solution

Given:

  $\text{I}\left(\text{x}\right)\text{=}\frac{{\text{11x}}^{\text{2}}\text{+33x+22}}{\text{x+2}}$ .

Concept used:

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:

  $\text{I}\left(\text{x}\right)\text{=}\frac{{\text{11x}}^{\text{2}}\text{+33x+22}}{\text{x+2}}$ .

Taking $\text{11}$ common from the numerator:

  $\text{I}\left(\text{x}\right)\text{=}\frac{\text{11}\left({\text{x}}^{\text{2}}\text{+3x+2}\right)}{\text{x+2}}$ .

Taking the middle term factorization on the denominator:

  $\text{I}\left(\text{x}\right)\text{=}\frac{\text{11}\left({\text{x}}^{\text{2}}\text{+2x+x+2}\right)}{\text{x+2}}$ .

  $\text{I}\left(\text{x}\right)\text{=}\frac{\text{11}\left(\text{x}\left(\text{x+2}\right)\text{+1}\left(\text{x+2}\right)\right)}{\text{x+2}}$ .

  $\text{I}\left(\text{x}\right)\text{=}\frac{\text{11}\left(\text{x+2}\right)\left(\text{x+1}\right)}{\text{x+2}}$ .

Dividing numerator and denominator both by $\text{x+2}$ :

  $\text{I}\left(\text{x}\right)\text{=11}\left(\text{x+1}\right)$ .

Hence, the function of average hourly income will be simplified to $\text{I}\left(\text{x}\right)\text{=11x+11}$ .