1. ## Polynomial

The Polynomial x^4 + 5x +a is denoted by p(x). It is Given that x² - x +3 is a factor of p(x).

a) Find the Value of a and factorize completely.

b) Hence state the number of real roots of the equation p(x) = 0 , Justifying your answer.

2. Originally Posted by Shannon
The Polynomial x^4 + 5x +a is denoted by p(x). It is Given that x² - x +3 is a factor of p(x).

a) Find the Value of a and factorize completely.

b) Hence state the number of real roots of the equation p(x) = 0 , Justifying your answer.
Hi Shannon,

Using polynomial long division, it can be determined that $a = -6$.

Divide $x^4+5x+a$ by $x^2-x+3$ to determine the value of a than makes the remainder equal to zero.

The factors of $p(x)=x^4+5x-6$ are $p(x)=(x^2-x+3)(x^2+x-2)\Longrightarrow p(x)=(x^2-x+3)(x+2)(x-1)$

There are two real zeros at -2 and 1.

There are two imaginary zeros when you solve $x^2-x+3=0$

3. Originally Posted by Shannon
The Polynomial x^4 + 5x +a is denoted by p(x). It is Given that x² - x +3 is a factor of p(x).

a) Find the Value of a and factorize completely.

b) Hence state the number of real roots of the equation p(x) = 0 , Justifying your answer.
Hi Shannon,

$x^4+5x+a=(x^2-x+3)(x^2+bx+c)=x^4+bx^3+cx^2-x^3-bx^2-cx+3x^2+3bx+3c$

$x^4+(0)x^3+(0)x^2+5x+a=x^4+(b-1)x^3+(c-b+3)x^2+(3b-c)x+3c$

$\Rightarrow\ b=1,\ \Rightarrow\ c=-2,\ \Rightarrow\ a=-6$

$x^4+5x-6=(x^2-x+3)(x^2+x-2)=(x^2-x+3)(x+2)(x-1)$

There are 2 real roots, as the roots of $x^2-x+3$ are

$x=\frac{1\pm\sqrt{1-4(3)}}{2}$

which are complex