Question

If [x2-1] is a factor of ax4+bx+cx2+dx+e shows that a+c+e=b+d=0

Answer 1

Question :-

If x² - 1 is a factor of ax⁴ + bx³ + cx² + dx + e . Show that a + c + e = b + d = 0

Answer :-

Given :-

x² - 1 is a factor of ax⁴ + bx³ + cx² + dx + e .

Required to find :-

• Show that a + c + e = b + d = 0

Solution :-

Given data :-

x² - 1 is a factor of ax⁴ + bx³ + cx² + dx + e

we need to show that a + c + e = b + d = 0 .

Let's consider the given polynomial as ;

p ( x ) = ax⁴ + bx³ + cx² + dx + e

x² - 1 is a factor of p ( x )

So,

Let

=> x² - 1 = 0

=> x² = 1

=> x = √1

=> x = + 1 or - 1

This implies ;

p ( 1 ) =

☛ a ( 1 )⁴ + b ( 1 )³ + c ( 1 )² + d ( 1 ) + e = 0

☛ a ( 1 ) + b ( 1 ) + c ( 1 ) + d + e = 0

☛a + b + c + d + e = 0 $\longrightarrow{\tt{\bf{Equation - 1 }}}$

consider this as equation - 1

Similarly,

☛ p ( - 1 ) =

☛ a ( - 1 )⁴ + b ( - 1 )³ + c ( - 1 )² + d ( - 1 ) + e = 0

☛ a ( 1 ) + b ( - 1 ) + c ( 1 ) - d + e = 0

☛ a - b + c - d + e = 0

☛ a + c + e = b + d $\longrightarrow{\tt{\bf{Equation - 2 }}}$

Consider this as equation - 2

According to the question ;

Consider equation - 1

☛ a + b + c + d + e = 0

☛ a + c + e + b + d = 0

Substitute the value of equation 2 in equation 1

☛ b + d + b + d = 0

☛ 2b + 2d = 0

Taking 2 common

☛ 2 ( b + d ) = 0

☛ b + d = 0/2

☛ b + d = 0

Hence,

b + d = 0

Therefore ,

a + c + e = b + d = 0

Additional information :-

There is a lot difference between √1 and √- 1

√1 = + 1 or - 1

whereas ;

√- 1 = + i or - i

Here i stands for iota .

This belongs to complex number system .

These are known as " Imaginary Numbers " .

Answer 2

$\huge\bf\red{SolutioN:}$

Since x2 - 1 = (x - 1) is a factor of

p(x) = ax4 + bx3 + cx2 + dx + e

∴ p(x) is divisible by (x+1) and (x-1) separately

⇒ p(1) = 0 and p(-1) = 0

p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0

⇒ a + b + c + d + e = 0 ---- (i)

Similarly, p(-1) = a (-1)4 + b (-1)3 + c (-1)2 + d (-1) + e = 0

⇒ a - b + c - d + e = 0

⇒ a + c + e = b + d ---- (ii)

$\mathbb\red{Putting \: the \: value \: of a + c + e \: in \:eqn , \:we \: get :}$

a + b + c + d + e = 0

⇒ a + c + e + b + d = 0

⇒ b + d + b + d = 0

⇒ 2(b+d) = 0

⇒ b + d = 0 ---- (iii)

comparing equations (ii) and (iii) , we get

$\mathsf{\fbox{\fbox{\green{a + c + e = b + d = 0}}}}$