Question

find the zero of the polynomial p(x)=x2-16​

Answer 1

Given -

• p(x) = x² - 16

To Find -

• Zeroes of the polynomial

Now,

→ x² - 16

→ (x)² - (4)²

• Using the identity of a² - b² = (a + b)(a - b)

→ (x + 4)(x - 4)

Zeroes are -

→ x + 4 = 0 and x - 4 = 0

→ x = -4 and x = 4

Verification :-

• α + β = -b/a

→ -4 + 4 = -(0)/1

→ 0 = 0

LHS = RHS

And

• αβ = c/a

→ -4 × 4 = -16/1

→ -16 = -16

LHS = RHS

Hence,

Verified...

It shows that our answer is absolutely correct.

Answer 2

$\large{\underline{\boxed{\text{Given:}}}}</p><p>$

$\text{p(x) = } {x}^{2} - 16$

$\large{\underline{\boxed{\text{To find:}}}}</p><p>$

• The zeroes of the polynomial

$\large{\underline{\boxed{\text{Solution:}}}}</p><p>$

p(x) = x² - 16

can be written as

=> (x)² - (4)²

Now, using identity

a² – b² = (a+b) (a-b)

=> (x+4) (x-4)

Zeroes :-

x +4 = 0 and x -4 = 0

x = -4 and x = 4

_____________________

Another way:–

p(x) = x² - 16

=> x² - 16 = 0

=> x² = 16

=> $\text{x} = \sqrt{16}$

=> $\text{x} = \pm{4}$

=> x = 4 and x = -4