Question

Please help, this is due in 20 minutes!

Answer 1

f(x) = -9

g(x) = 27

Explanation:

Given rational expression:

`(18x)/(8x^2-10x+3)`

To express the given rational expression in partial fractions, first we need to factor the denominator:

`\begin{aligned}8x^2-10x+3&=8x^2-6x-4x+3\\&=2x(4x-3)-1(4x-3)\\&=(2x-1)(4x-3)\end{aligned}`

Therefore:

`(18x)/((2x-1)(4x-3))`

Now write as an identity with partial fractions:

`(18x)/((2x-1)(4x-3)) \equiv (A)/(2x-1)+(B)/(4x-3)`

Add the partial fractions:

`\begin{aligned}(18x)/((2x-1)(4x-3)) &\equiv (A(4x-3))/((2x-1)(4x-3))+(B(2x-1))/((2x-1)(4x-3))\\\\&\equiv (A(4x-3)+B(2x-1))/((2x-1)(4x-3))\end{aligned}`

Cancel the denominators from both sides of the original identity, so the numerators are equal:

`18x \equiv A(4x-3)+B(2x-1)`

To solve by the method of substitution, we can substitute values of x which make one of the expressions in brackets equal zero to get rid of all but one of A and B.

Substituting x = 1/2:

`\begin{aligned}x=(1)/(2) \implies 18\left((1)/(2)\right)& =A\left(4\left((1)/(2)\right)-3\right)+B\left(2\left((1)/(2)\right)-1\right)\\\\9 &=A\left(-1\right)+B\left(0\right)\\\\A&=-9\end{aligned}`

Substituting x = 3/4:

`\begin{aligned}x=(3)/(4) \implies 18\left((3)/(4)\right)& =A\left(4\left((3)/(4)\right)-3\right)+B\left(2\left((3)/(4)\right)-1\right)\\\\(27)/(2)&=A\left(0\right)+B\left((1)/(2)\right)\\\\B&=27\end{aligned}`

Finally, replace A and B in the original identity:

`(18x)/((2x-1)(4x-3)) \equiv (-9)/(2x-1)+(27)/(4x-3)`

Therefore:

• 
  `f(x) = -9`

• 
  `g(x) = 27`