The remainder when x^2 - 3x + 1 is divided by (x + d) is 11. Find the possible values of d.
Since the divisor is x + d, then the zero (when doing the synthetic division) is -d. So do the synthetic division:
Set the remainder equal to the given value:Code:-d | 1 -3 1 | -d d^2+3d +------------------ 1 -d-3 d^2+3d+1
. . . . .$\displaystyle d^2\, +\, 3d\, +\, 1\, =\, 11$
This rearranges as:
. . . . .$\displaystyle d^2\, +\, 3d\, -\, 10\, =\, 0$
Either factor the quadratic and then solve the factors for the values of d, or else plug the above into the Quadratic Formula.
Hello, Mr Rayon!
From the heading, I assume you're expected to know the Remainder Theorem.The remainder when $\displaystyle x^2 - 3x + 1$ is divided by $\displaystyle (x + d)$ is 11.
Find the possible values of $\displaystyle d.$
. . If a polynomial $\displaystyle p(x)$ is divided by $\displaystyle (x-a)$, the remainder is $\displaystyle p(a).$
We have: .$\displaystyle p(x) \:=\:x^2-3x+1$
When $\displaystyle p(x)$ is divided by $\displaystyle (x+d)$, the remainder is $\displaystyle p(\text{-}d)$
We have: .$\displaystyle p(x) \:=\:x^2-3x+1$ divided by $\displaystyle (x-[\text{-}d])$
Hence: .$\displaystyle (\text{-}d)^2 - 3(\text{-}d) + 1 \:=\:1 \quad\Rightarrow\quad d^2 + 3d \:=\:0 \quad\Rightarrow\quad d(d+3) \:=\:0$
Therefore: .$\displaystyle d \;=\;0,\:\text{-}3$