Find the number of real zeros of the followin | vedclass

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

Question

(D) To find the zeros of the polynomial $p(x) = x^{3} - 9x$,we set $p(x) = 0$.$x^{3} - 9x = 0$Factor out $x$ from the expression:$x(x^{2} - 9) = 0$Usi

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) To find the zeros of the polynomial $p(x) = x^{3} - 9x$,we set $p(x) = 0$.$x^{3} - 9x = 0$Factor out $x$ from the expression:$x(x^{2} - 9) = 0$Using the difference of squares formula $a^{2} - b^{2} = (a - b)(a + b)$,we can factor $x^{2} - 9$ as $(x - 3)(x + 3)$:$x(x - 3)(x + 3) = 0$Setting each factor to zero,we get:$x = 0$,$x - 3 = 0 \implies x = 3$,and $x + 3 = 0 \implies x = -3$.Thus,the zeros of the polynomial are $0, 3, -3$.Since there are $3$ distinct real values,the number of real zeros is $3$.

Find the number of real zeros of the following polynomial: $p(x) = x^{3} - 9x$.

Similar Questions

The zeros of $p(x) = 2 + x - x^{2}$ are.......

If $P(-7) = 0$,then one factor of $P(x)$ is . . . . . . .

Obtain a quadratic polynomial with the following conditions:
The sum of the zeros $= -\frac{1}{4}$;
The product of the zeros $= \frac{1}{4}$.

The zeros of $p(x) = x^{2} + 4x + 3$ are.........

Answer the following and justify:
If on division of a polynomial $p(x)$ by a polynomial $g(x)$,the quotient is zero,what is the relation between the degrees of $p(x)$ and $g(x)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module