Question

tje value of p(x) =(x-1)(×+1) for p(1) is​

Answer 1

GIVEN :-

• p(x) = (x - 1)(x + 1)

TO FIND :-

• The value of p(1)

SOLUTION :-

➳ p(x) = (x - 1)(x + 1)

As we know the algebraic identity :- (a + b)(a - b) = a² - b². So let's apply this identity,

➳ p(x) = x² - 1²

➳ p(1) = (1)² - (1)²

➳ p(1) = 1 - 1

➳ p(1) = 0

Hence value of the given polynomial p(x) at p(1) is 0.

ADDITIONAL INFORMATION :-

Algebric Identities :

$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\1)\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$

Answer 2

Question :

The value of p ( x ) = ( x - 1 ) ( x + 1 )

Answer :

Given :

• p ( x ) = ( x - 1 ) ( x + 1 )

To find :

• The value of p ( 1 )

According to the question :

⇢p ( x ) = ( x - 1 ) ( x + 1 )

⇢p ( x ) = x² - 1

This is in the Form,

➳ [ ( a + b ) ( a - b ) = a² - b² ]

↦p ( x ) = p ( 1 ) { given }

↦x = 1

Substituting ' x ' value :

⟹ p ( x ) = x² - 1

⟹ p ( 1 ) = ( 1 )²- 1

⟹ p ( 1 ) = ( 1 × 1 ) - 1

⟹ p ( 1 ) = 1 - 1

⟹ p ( 1 ) = 0

∴ p ( 1 ) = 0

So, It's Done !!