Question

find the zeros of polynomial p ( x)=x4-16

Answer 1

for zeroes of the polynomial,

p(x)=0,

x⁴-16=0,

(x²)²-(2²)²=0,

(x²-2²)(x²+2²)=0,

(x²-4)(x²+4)=0,

if

x²-4=0,

then

x²-4=0,

x²=4,

x=√4,

x=+-2,


if

x²+4=0,

then

x²=-4,

x=√(-4),

x=2i

Answer 2

The zeros of polynomial p(x) = x⁴ - 16 are - 2 , 2 , - 2i 2i

Given :

The polynomial p(x) = x⁴ - 16

To find :

The zeroes of the polynomial

Solution :

Step 1 of 2 :

Write down the given polynomial

Here the given polynomial is p(x) = x⁴ - 16

Step 2 of 2 :

Find the zeroes of the polynomial

For zeroes of the polynomial p(x) we have

$\displaystyle \sf{ p(x) = 0 }$

$\displaystyle \sf{ \implies {x}^{4} - 16 = 0}$

$\displaystyle \sf{ \implies {( {x}^{2}) }^{2} - {4}^{2} = 0}$

$\displaystyle \sf{ \implies ( {x}^{2} + 4)( {x}^{2} - 4) = 0 }\: \: \: \bigg[ \: \because \: {a}^{2} - {b}^{2} = (a + b)(a - b) \bigg]$

$\displaystyle \sf{ \implies [ {(x)}^{2} - {(2i)}^{2} ]( {x}^{2} - {2}^{2} ) = 0 }$

$\displaystyle \sf\implies (x + 2i)(x - 2i)(x + 2)(x - 2)= 0 \: \: \: \bigg[ \: \because \: {a}^{2} - {b}^{2} = (a + b)(a - b) \bigg]$

Now ,

x + 2i = 0 gives x = - 2i

x - 2i = 0 gives x = 2i

x + 2 = 0 gives x = - 2

x - 2 = 0 gives x = 2

Hence the zeros of polynomial p(x) = x⁴ - 16 are - 2 , 2 , - 2i 2i

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