Rational Expressions of Polynominals

There are constants  $\alpha$ and $\beta$ such that $\dfrac{x-\alpha}{x+\beta} = \dfrac{x^2-80x+1551}{x^2+57x-2970}$.
What is $\alpha+\beta$ ?

1.

$\begin{array}{|rcll|} \hline && x^2-80x+1551 \\ &=& (x-40)^2 - 40^2+1551 \\ &=& (x-40)^2 - 1600+1551 \\ &=& (x-40)^2 - 49 \\ &=& (x-40)^2 - 7^2 \\ &=& [(x-40)-7][(x-40)+7] \\ &=& (x-47)(x-33) \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline && x^2+57x-2970 \\ &=& \left( x+\frac{57}{2} \right)^2 - \left(\frac{57}{2}\right)^2 -2970 \\ &=& \left( x+28.5 \right)^2 - 28.5^2 -2970 \\ &=& \left( x+28.5 \right)^2 - 3782.25 \\ &=& \left( x+28.5 \right)^2 - 61.5^2 \\ &=& [(x+28.5)-61.5][(x+28.5)+61.5] \\ &=& (x-33)(x+90) \\ \hline \end{array}$

3.

$\begin{array}{|rcll|} \hline && \dfrac{x^2-80x+1551}{x^2+57x-2970} \\\\ &=& \dfrac{(x-47)(x-33)}{(x-33)(x+90)} \\\\ &=& \dfrac{x-47}{x+90} \\ \hline \end{array}$

4.

$\begin{array}{|rcll|} \hline \dfrac{x-\alpha}{x+\beta} &=& \dfrac{x-47}{x+90} \\ \\ \hline x-\alpha &=& x-47 \\ -\alpha &=& -47 \\ \mathbf{\alpha} & \mathbf{=} & \mathbf{47} \\ \\ \hline x+\beta &=& x+90 \\ \mathbf{\beta} & \mathbf{=} & \mathbf{90} \\ \\ \hline \alpha + \beta &=& 47+90 \\ &=& 137 \\ \hline \end{array}$

$\mathbf{\alpha+\beta =137}$

Since the left hand side is the ratio of two linear terms the right hand side must be too. In other words the numerator and denominator on the right hand side must contain a common factor, say $(x-\gamma)$ which cancels out. The right hand side can therefore be written:

$\displaystyle{(x-\alpha)(x-\gamma)\over(x+\beta)(x-\gamma)} ={x^2-(\alpha+\gamma)x+\alpha\gamma\over x^2+(\beta-\gamma)x -\beta\gamma} ={x^2-80x+1551\over x^2+57x-2970}$,

so equating coefficients of x in the polynomials we must have $\alpha+\gamma=80$ and $\beta-\gamma=57$.

Adding these two expressions together $\gamma$ cancels and we have immediately:

$\alpha+\beta=80+57=137$.