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 $\displaystyle a = -6$.
Divide $\displaystyle x^4+5x+a$ by $\displaystyle x^2-x+3$ to determine the value of a than makes the remainder equal to zero.
The factors of $\displaystyle p(x)=x^4+5x-6$ are $\displaystyle p(x)=(x^2-x+3)(x^2+x-2)\Longrightarrow p(x)=(x^2-x+3)(x+2)(x-1)$
There are two real zeros at -2 and 1.
There are two imaginary zeros when you solve $\displaystyle x^2-x+3=0$
Hi Shannon,
$\displaystyle x^4+5x+a=(x^2-x+3)(x^2+bx+c)=x^4+bx^3+cx^2-x^3-bx^2-cx+3x^2+3bx+3c$
$\displaystyle x^4+(0)x^3+(0)x^2+5x+a=x^4+(b-1)x^3+(c-b+3)x^2+(3b-c)x+3c$
$\displaystyle \Rightarrow\ b=1,\ \Rightarrow\ c=-2,\ \Rightarrow\ a=-6$
$\displaystyle x^4+5x-6=(x^2-x+3)(x^2+x-2)=(x^2-x+3)(x+2)(x-1)$
There are 2 real roots, as the roots of $\displaystyle x^2-x+3$ are
$\displaystyle x=\frac{1\pm\sqrt{1-4(3)}}{2}$
which are complex