Question

Please help, this is due in 20 mins

Answer 1

f(x) = 4

g(x) = -3x+2

Explanation:

Given rational expression:

`(x^2-21x+26)/((x+5)(x^2-2x+4))`

As the denominator has a linear factor and irreducible quadratic factor, the partial fraction form is:

`\boxed{(N(x))/((ax+b)(x^2+bx+c)) \equiv (A)/(ax+b)+(Bx+C)/(x^2+bx+c)}`

Therefore, the given rational expression as an identity with partial fractions is:

`(x^2-21x+26)/((x+5)(x^2-2x+4))\equiv (A)/(x+5)+(Bx+C)/(x^2-2x+4)`

Add the partial fractions:

`\begin{aligned}(x^2-21x+26)/((x+5)(x^2-2x+4))&\equiv (A(x^2-2x+4))/((x+5)(x^2-2x+4))+((Bx+C)(x+5))/((x+5)(x^2-2x+4))\\\\&\equiv (A(x^2-2x+4)+(Bx+C)(x+5))/((x+5)(x^2-2x+4))\end{aligned}`

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

`x^2-21x+26\equiv A(x^2-2x+4)+(Bx+C)(x+5)`

Expand the right side:

`\begin{aligned}x^2-21x+26&\equiv A(x^2-2x+4)+(Bx+C)(x+5)\\\\&\equiv Ax^2-2Ax+4A+Bx^2+5Bx+Cx+5C\\\\&\equiv Ax^2+Bx^2+5Bx-2Ax+Cx+4A+5C\\\\&\equiv (A+B)x^2+(5B-2A+C)x+(4A+5C)\end{aligned}`

Compare the coefficients:

`\begin{aligned}&\textsf{Equating $x^2$ coefficients:}&\quad 1&=A+B\\\\&\textsf{Equating $x$ coefficients:}&\quad -21&=5B-2A+C\\\\&\textsf{Equating constant terms:}&\quad 26&=4A+5C\end{aligned}`

Solving these equations simultaneously gives:

`A=4`

`B=-3`

`C=2`

Finally, replace A, B and C in the original identity:

`(x^2-21x+26)/((x+5)(x^2-2x+4))\equiv (4)/(x+5)+(-3x+2)/(x^2-2x+4)`

Therefore:

• 
  `f(x) = 4`

• 
  `g(x) = -3x+2`