The remainder when x^2 - 3x + 1 is divided by (x + d) is 11. Find the possible values of d.

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- Apr 7th 2009, 07:53 AMMr RayonThe remainder and factor theorems
The remainder when x^2 - 3x + 1 is divided by (x + d) is 11. Find the possible values of d.

- Apr 7th 2009, 08:22 AMstapel
Since the divisor is x + d, then the zero (when doing the

**synthetic division**) is -d. So do the synthetic division:

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**.

(Wink) - Apr 7th 2009, 08:58 AMSoroban
Hello, Mr Rayon!

Quote:

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: .$\displaystyle d \;=\;0,\:\text{-}3$

- Apr 7th 2009, 04:15 PMMr Rayon
- Apr 7th 2009, 04:29 PMmr fantastic
- Apr 7th 2009, 04:40 PMMr Rayon
- Apr 7th 2009, 04:41 PMstapel
- Apr 7th 2009, 04:45 PMMr Rayon