Question

(iv) p(x) = (x - 1)(x+1)​

Answer 1

Step-by-step explanation:

Hey mate !!

Here's your answer !!

1 and 2 are zeros of polynomial means that if the value is substituted for x then the equation turns zero.

So let's prove using x = 1 first

\begin{gathered}= (x - 1)(x - 2) \\ = (1 - 1)(1 - 2) \\ = (0)( - 1) \\ = 0\end{gathered}=(x−1)(x−2)=(1−1)(1−2)=(0)(−1)=0

Since the equation turns to be zero, 1 is the zero of the polynomial.

Now let's take x = 2

\begin{gathered}= (x - 1)(x - 2) \\ = (2 - 1)(2 - 2) \\ = (1)(0) \\ = 0\end{gathered}=(x−1)(x−2)=(2−1)(2−2)=(1)(0)=0

Since the equation turns to be zero, 2 is also the zero of the polynomial.

Hence Verified !!

Hope it helps !!

Answer 2

Answer:

p(x) = x² - 1²

= x²- 1

Step-by-step explanation:

May be it helps you dear friend.