Question

Find the other two zeroes of the polynomial x4 - 5x + 2x2 + 10x - 8, if it Is given that the two zeroes are - √2 and √2​

Answer 1

Answer:

$\large\boxed{\{-\sqrt2,\ 1,\ \sqrt2,\ 4\}}$

Step-by-step explanation:

$W(x)=x^4 - 5x^3 + 2x^2 + 10x - 8$

Let's look for zeros in constant term dividers.

List of 8 divisors:

$\{\pm1,\ \pm2,\ \pm4,\ \pm8\}$

Check:

$W(1)=1^4-5(1)^3+2(1)^2+10(1)-8=1-5(1)+2(1)+10\\\\=1-5+2+10-8=0\\\\W(-1)=(-1)^4-5(-1)^3+2(-1)^2+10(-1)-8=1-5(-1)+2(1)-10-8\\\\=1+5+2-10-8=-10\neq0\\\\W(2)=2^4-5(2)^3+2(2)^2+10(2)-8=16-5(8)+2(4)+20-8\\\\=16-40+8+20-8=-4\neq0\\\\W(-2)=(-2)^4-5(-2)^3+2(-2)^2+10(-2)-8\\\\=16-5(-8)+2(4)-20-8=16+40+8-20-8=36\neq0\\\\W(4)=4^4-5(4)^3+2(4)^2+10(4)-8=256-5(64)+2(16)+48-8\\\\=256-320+32+40-8=0$

We found missing zeros. Therefore, we do not check the remaining numbers.