1. ## Polynomial

Question>> the polynomial x^4+5x+a is denoted by p(x). It is given that x^2-x+3 is a factor of p(x).
part (1) Find the value of a in p(x).

2. ## Re: Polynomial

$(x^2 - x + 3)(a_2x^2 + a_1x + a_0) = x^4 + 5x + a$ so [tex] we get $a_2x^4 + a_1x^3 + a_0x^2 - a_2x^3 - a_1 x^2 - a_0x + a_2*3*x^2 + 3*a_1x + 3a_0= a_2x^4 + (a_1 - a_2)x^3 + (a_0 - a_1 + 3a_2)x^2 + (3a_1 - a_0)x + 3a_0 = x^4 + 5x + a$ which means a_2 = 1. So $a_2x^4 + (a_1 - 1)x^3 + (a_0 - a_1 + 3)x^2 + (3a_1 - a_0)x + 3a_0$ and since $a_1 - 1 = 0$ $a_1 = 1$. So now you got $a_2x^4 + (a_0 + 2)x^2 + (3 - a_0)x + 3a_0$ and since $a_0 + 2 = 0$ $a_0 = -2$ so again $x^4 + (5)x -6$ so a = -6.

3. ## Re: Polynomial

Code:
................. x^2  +  x  - 2
.................-----------------------------
x^2 - x + 3 | x^4 + 0x^3 + 0x^2 + 5x + a
............  x^4 -  x^3  + 3x^2
..............-----------------------------
....................x^3  - 3x^2 + 5x + a
....................x^3  -  x^2 + 3x
....................---------------------
........................- 2x^2 + 2x + a
........................- 2x^2 + 2x - 6
........................----------------
......................................a+6
remainder, a+6 = 0

a = -6

4. ## Re: Polynomial

thanks jakncoke. Thankyou skeeter, i was also using the same method but my remainder included the variable 'x' so i thought it was wrong. Found my mistake though, thanks once again.