Question

If two zeroes of the polynomial x4+3x3-20x2-6x+36 are root2 and -root2 find the other zeroes of the polynomial

Answer 1

Answer:

3 and -6

Step-by-step explanation:

$p(x) = x^4 + 3x^3 - 20x^2 - 6x + 36$

√2 and -√2 are two zeroes of p(x).

∴ (x - √2), (x + √2) are two factors of p(x).

Divide p(x) by (x - √2).

$x^4+3x^3-20x^2-6x+36 \\ \\ x^4-\sqrt{2}x^3+(3+\sqrt{2})x^3-(3\sqrt{2}+2)x^2-(18-3\sqrt{2})x^2+(18\sqrt{2}-6)x-18\sqrt{2}x + 36 \\ \\ x^3(x-\sqrt{2})+x^2(3+\sqrt{2})(x-\sqrt{2})-x(18-3\sqrt{2})(x-\sqrt{2})-18\sqrt{2}(x-\sqrt{2}) \\ \\ (x-\sqrt{2})(x^3+(3+\sqrt{2})x^2-(18-3\sqrt{2})x-18\sqrt{2})$

Let the quotient  x³ + (3 + √2)x² - (18 - 3√2)x - 18√2  be q(x).

Divide q(x) by (x + √2).

$x^3+(3+\sqrt{2})x^2-(18-3\sqrt{2})x-18\sqrt{2} \\ \\ x^3+\sqrt{2}x^2+3x^2+3\sqrt{2}x-18x-18\sqrt{2} \\ \\ x^2(x+\sqrt{2})+3x(x+\sqrt{2})-18(x+\sqrt{2}) \\ \\ (x+\sqrt{2})(x^2+3x-18)$

Let the quotient  x² + 3x - 18  be r(x).

Zeroes of r(x) are the answers. Let's find it.

$x^2+3x-18 \\ \\ x^2-3x+6x-18 \\ \\ x(x-3)+6(x-3) \\ \\ (x-3)(x+6)$

∴ The answers are 3 and -6.

Hope this may be helpful.

Please mark my answer as the brainliest if this may be helpful.

Thank you. Have a nice day. :-)

#adithyasajeevan