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).
Can somebody please reply to this quickly. Thank you
$\displaystyle (x^2 - x + 3)(a_2x^2 + a_1x + a_0) = x^4 + 5x + a$ so [tex] we get $\displaystyle a_2x^4 + a_1x^3 + a_0x^2 - a_2x^3 - a_1 x^2 - a_0x + a_2*3*x^2 + 3*a_1x + 3a_0= a_2x^4 + (a_1 - a_2)x^3 + (a_0 - a_1 + 3a_2)x^2 + (3a_1 - a_0)x + 3a_0 = x^4 + 5x + a$ which means a_2 = 1. So $\displaystyle a_2x^4 + (a_1 - 1)x^3 + (a_0 - a_1 + 3)x^2 + (3a_1 - a_0)x + 3a_0$ and since $\displaystyle a_1 - 1 = 0 $ $\displaystyle a_1 = 1 $. So now you got $\displaystyle a_2x^4 + (a_0 + 2)x^2 + (3 - a_0)x + 3a_0$ and since $\displaystyle a_0 + 2 = 0 $ $\displaystyle a_0 = -2 $ so again $\displaystyle x^4 + (5)x -6$ so a = -6.
remainder, a+6 = 0Code:................. 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
a = -6