Question

Help please with this question

Answer 1

Given:

The given two functions are $f(x)=\frac{4 x^{2}+24 x}{x^{2}-11 x+30}$  and  $g(x)=\frac{x-5}{x^{2}}$

We need to determine the value of R(x)

Value of R(x):

The value of R(x) can be determined by R(x) = f(x) × g(x)

Substituting the values, we get;

$R(x)=\frac{4 x^{2}+24 x}{x^{2}-11 x+30} \cdot \frac{x-5}{x^{2}}$

Multiplying the fractions, we have;

$R(x)=\frac{\left(4 x^{2}+24 x\right)(x-5)}{\left(x^{2}-11 x+30\right) x^{2}}$

Let us factor out the common term 4x from the term (4x² + 24x)

Thus, we have;

$R(x)=\frac{4 x(x+6)(x-5)}{\left(x^{2}-11 x+30\right) x^{2}}$

Let us factor the term (x² -11x + 30), we get;

$R(x)=\frac{4(x+6)(x-5)}{x(x-5)(x-6)}$

Cancelling the common term, we have;

$R(x)=\frac{4(x+6)}{x(x-6)}$

Thus, the value of R(x) is $\frac{4(x+6)}{x(x-6)}$