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),?

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),
Find the value of a and factorize p(x) completely.

Answer 1

x² - x + 3 is a factor thus remainder = 0 upon division
Set up the division:-
____________x² + x - 2
x² - x + 3___|x^4 + 5x + a
__________|x^4 - x³ + 3x²
__________|x³ - 3x² + 5x + a
__________|x³ - x² + 3x
__________|- 2x² + 2x + a
__________|- 2x² + 2x - 6
__________|a + 6

R = a + 6 = 0
a = - 6

Factors are:-
(x² - x + 3)(x² + x - 2)
(x² - x + 3) (x + 2)(x - 1)

Answer 2

Since x^2-x+3 is a factor of P(x), therefore it must dividie exactly into it. So if you use algebraic long division, you get
P(x)= Quotient(x)Divider(x) + Remainder
= (x^2-x+3)(x^2+x-2) + (a+6)
And as it is a factor the remainder must equal 0.
Therefore:
a+6=0
a=-6
Therefore:
P(x)=(x^2-x+3)(x^2+x-2) and a=-6

Answer 3

Since x^2 - x + 3 is a factor of x^4+5x+a , therefore we can say that remainder should be zero on dividing the polynomial by the factor.
x^2 - x + 3 ) x^4+5x+a (x^2+x-2
x^4-x^3+3X^2+5x+a
------------------------------
x^3-3x^2+5x+a
x^3 - x^2+3x
-------------------------------
-2x^2+2x+a
-2x^2+2x-6
------------------------------
a+6
therefore ,a+6=0
=> a= -6
hence:(x^4+5x+a)=(x^2-x+3)(x^2+x-2)
To factorize it further it will be in complex form.so you can leave it till above.