The zeros of p(x) = x^{3} - 4x are .......... | 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 $p(x) = x^{3} - 4x$,we set $p(x) = 0$.$x^{3} - 4x = 0$Factor out $x$ from the expression:$x(x^{2} - 4) = 0$Usi

Vedclass · Accepted Answer

$0, \pm 2$

Vedclass · Answer

(D) To find the zeros of the polynomial $p(x) = x^{3} - 4x$,we set $p(x) = 0$.$x^{3} - 4x = 0$Factor out $x$ from the expression:$x(x^{2} - 4) = 0$Using the difference of squares formula $a^{2} - b^{2} = (a - b)(a + b)$,we can write $x^{2} - 4$ as $(x - 2)(x + 2)$:$x(x - 2)(x + 2) = 0$Setting each factor to zero,we get:$x = 0$,$x - 2 = 0 \Rightarrow x = 2$,and $x + 2 = 0 \Rightarrow x = -2$.Thus,the zeros are $0, 2, -2$,which can be written as $0, \pm 2$.

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

Similar Questions

For the given sum and product of zeroes,find the quadratic polynomial. Also,find the zeroes of this polynomial by factorization:
Sum of zeroes = $-2 \sqrt{3}$,Product of zeroes = $-9$.

If $\alpha, \beta$ and $\gamma$ are the zeros of the cubic polynomial $p(x) = x^{3} - 3x^{2} - 6x + 8$,then $\alpha \beta \gamma = \dots$

Find the value of $p(x) = 2x^3 + x^2 - x - 1$ at $x = 0$ and $x = -2$.

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

Identify the type of the given polynomial based on its degree: $p(x) = (x - 1)(2 - x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module