How do you evaluate #f(x)=2x^3-x^4+5x^2-x# at x=3 using direct substitution and synthetic division?
1 Answer
Explanation:
Direct substitution is probably the easiest to understand. All we're doing is plugging in the
If
#f(x)=2x^3-x^4+5x^2-x#
then
#f(color(blue)3)=2(color(blue)3)^3-(color(blue)3)^4+5(color(blue)3)^2-color(blue)3#
#color(white)(f(3))=2(27)-" "81" "+" "5(9)" "-3#
#color(white)(f(3))=54-81+45-3#
#color(white)(f(3))=15#
To use synthetic division, we are going to "factor out"
First: arrange the coefficients of the polynomial in decreasing order in an "L" frame, including
Next, attempt to factor out
The remainder of
Bonus:
I'll admit, I don't remember learning to evaluate polynomials with synthetic division when I was in high school, but I can see why this works. The above synthetic division yields the following "factored" version of
#f(x)=(x-3)("-"x^3-x^2+2x+5+15/(x-3))#
This can be (partially) re-distributed as:
#f(x)=(x-3)("-"x^3-x^2+2x+5)+(15(x-3))/(x-3)#
#color(white)(f(x))=(x-3)("-"x^3-x^2+2x+5)+15#
(Cancelling the
From here, it is easy to see that when
#f(color(red)3)=(color(red)3-3)(..." "..." "..." "...)+15#
#color(white)(f(3))=(" "0" ")(..." "..." "..." "...)+15#
#color(white)(f(3))=15# .