Question

which are the zeros of p (x) = x^2 - 1​

Answer 1

Answer:

p(0)=(0)^2-1

p(0)=0-1

p(0)=-1

Answer 2

We have

$p(x) = {x}^{2} - 1$

We need to find the zeroes. First we will simply simplify it

So, ${x}^{2} = 1$

Now, square rooting both sides

We will get

$x = \sqrt{1}$

Since the value of $\sqrt{1}$ is 1 only.

Hence,

$\underline{\boxed { { \red{x = \pm1 }}}}$

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Verification :

${(1)}^{2} - 1 = 1 - 1 = 0$

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Another method:

This one is by factorising

We can write 1 as 1² also

$\implies \: p(x) = {x}^{2} - {1}^{2}$

Using identity

$\text{ a} ^{2} - \text{b}^{2} = \text{(a - b)(a + b)}$

We get

$p(x) = (x - 1)(x + 1)$

Hence, (x–1) and (x+1) are the factors.

Now,

$x - 1 = 0 \text{ \: and \: } x + 1 = 0 \\ \underline {\boxed{ \red{x = 1}}} \text{ \: and \: } \underline{ \boxed{ \red{x = - 1}}}$