Find the zeroes of the polynomial:p(x)=(x-2)^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To find the zeroes of the polynomial $p(x)$,we set $p(x) = 0$.Given $p(x) = (x-2)^{2} - (x+2)^{2}$.Using the algebraic identity $a^2 - b^2 = (a-b)

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) To find the zeroes of the polynomial $p(x)$,we set $p(x) = 0$.Given $p(x) = (x-2)^{2} - (x+2)^{2}$.Using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$,where $a = (x-2)$ and $b = (x+2)$:$p(x) = [(x-2) - (x+2)] \cdot [(x-2) + (x+2)] = 0$$p(x) = (x - 2 - x - 2) \cdot (x - 2 + x + 2) = 0$$p(x) = (-4) \cdot (2x) = 0$$-8x = 0$$x = 0$Thus,the zero of the polynomial is $0$.

Find the zeroes of the polynomial:
$p(x)=(x-2)^{2}-(x+2)^{2}$

Similar Questions

Show that $2x - 3$ is a factor of $x + 2x^3 - 9x^2 + 12$.

If $x+3$ is a factor of $x^{3}+12x^{2}+ax+60$,then $a = \ldots$

Is the following expression a polynomial? Justify your answer:
$\sqrt{3} x^{2}-2 x$

Find the zero of the polynomial in the following case:
$q(y) = \pi y + 3.14$

Classify the following as a constant,linear,quadratic,or cubic polynomial:
$1+x+x^{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the zeroes of the polynomial:$p(x)=(x-2)^{2}-(x+2)^{2}$