Question

When using the inspection method to rewrite the rational expression x^2-10x+14/x-8 , what constant term must the numerator have for to be one of its factors? What number must you add to (and subtract from) the numerator's constant term to derive (x-8) as a factor?
For the numerator to have (x-8)
as a factor, its constant term must be . To get this number, add and subtract from the numerator’s constant term.

Answer 1

General idea:

When we factor a trinomial of the form $x^2+bx+c$, we need two numbers m and n to write it as product of two binomials like$(x+m)(x+n)$, where $m +n = b \;\; and \;\; m\cdot n = c$.

In order to get two numbers we need to list the factor pairs of c and see which two factors add up to give b.

Applying the concept:

$\frac{x^2-10x+14}{x-8}$

We need to make alterations to the constant term in such a way that below equation is true.

$x^2-10x+c=(x-8)(x+n)\\ \\ Comparing \; x^2+bx+c=(x+m)(x+n) \; with \; above \; equation \;,\\\\ We\; get\; m = -8\; and \; b = -10\\\\ Also \; we \; know, \; m + n = b \; \; and \;\; m \cdot n= c\\ \\ -8+n=-10 \; \ [ Adding \; 8 \; on \; both \; sides]\\ \\ -8+n+8=-10+8 \; [Combining \; like \; terms \; on \; either \; side]\\ \\ n=-2 \\ \\ c=m \cdot n = -8 \cdot -2 = 16$

Conclusion:

Right now the numerator is $x^2 - 10x + 14$, we need to add 2 to the constant of the numerator expression to get $x^2 - 10x + 16$.

For the numerator to have (x-8) as a factor, its constant term must be 16 .