Question

Find all the zeroes of ${\displaystyle (x^4+x^3–23x^2–3x+60)}$, if it is given that two of its zeroes are ${\displaystyle \sqrt{3}\text{and}–\sqrt{3}}$. 

Answer 1

Let f(x) =${\displaystyle x^4+x^3–23x^2–3x+60}$
Since ${\displaystyle \sqrt{3}}$ and ${\displaystyle –\sqrt{3}}$ are the zeroes of f(x), it follows that each one of (x – √3) and (x + √3) is a factor of f(x).
Consequently, ${\displaystyle (x–\sqrt{3})(x+\sqrt{3})=(x^2–3)}$is a factor of f(x).
On dividing f(x) by ${\displaystyle (x^2–3)}$, we get:  

  

f(x) = 0
⇒ ${\displaystyle (x^2+x–20)(x^2–3)=0}$
⇒ ${\displaystyle (x^2+5x–4x–20)(x^2–3)}$
⇒ ${\displaystyle [x(x+5)–4(x+5)](x^2–3)}$
⇒ ${\displaystyle (x–4)(x+5)(x–\sqrt{3})(x+\sqrt{3})=0}$
⇒ ${\displaystyle x=4\text{or}x=-5\text{or}x=\sqrt{3}\text{or}x=-\sqrt{3}}$
Hence, all the zeroes are √3, -√3, 4 and -5.