Question

Find the zeroes of polynomial
p(x)=x⁴-16​

Answer 1

Answer:

Step-by-step explanation:

For a polynomial P (x), real number k is said to be zero of polynomial P (x), if P (k) = 0. Therefore, 1 and 2 are the zeros of polynomial x2 – 3x + 2. We have discussed polynomials and their zeros here.

Answer 2

Answer:

$p(x) = {x}^{4} - 16 \\ \\ = ({x}^{2} ) ^{2} - {4}^{2} \\ \\ = ( {x}^{2} - 4)( {x}^{2} + 4) \\ \\either \: {x}^{2} - 4 = 0 \\ \\ = > {x}^{2} = 4 \\ \\ = > x = \sqrt{4} \\ \\ = > x = 2 \\ \\ or \: {x}^{2} + 4 = 0 \\ \\ = > {x}^{2} = - 4 \\ \\ = > x = \sqrt{ - 4} \\ \\ = > x = - 2 \\ \\ formula \: used \\ \\ {a}^{2} - {b}^{2} \\ \\ = (a + b)(a - b)$

hope it helps uH