Question

FIND the VALUES A and B so that(x+1) and (x-1) are factors of *
x 4 + ax3 – 3x3 + 2x + b​

Answer 1

Question :-

$\textsf{Find the values of "a" and "b"}$ .

$\tt{\small{Such\; that \;(x+1) \;and\; (x-1) \;are\; the\; factors\; of \;{p(x) = {x}^{4} + a{x}^{3} - 3{x}^{3} +2x + b}}}$

$\rule{400}{4}$

Given :

$\mathsf{ p(x) = {x}^{4} +a{x}^{3} - 3{x}^{3} +2x + b}$

Let's take (x+1) & (x-1) as the factors .

$\rule{400}{4}$

Required to find :-

• Values of a and b ?

$\rule{400}{4}$

Solution :

Given expression :-

$\mathsf{ p(x) = {x}^{4} +a{x}^{3} - 3{x}^{3} +2x + b}$

As ,

(x+1) & (x-1) are the factors

Let ,

x + 1 = 0

x = -1

Now substitute this value in place of x in the above expression

$\tt{p(-1) = {(-1)}^{4} + a{(-1)}^{3} - 3{(-1)}^{3} + 2(-1) + b}$

$\tt{ p(-1) = 1 + a(-1) -3(-1) -2 + b }$

$\tt{ p(-1) = 1 - a + 3 - 2 + b}$

$\tt{ p(-1) = 4 - a -2 + b }$

$\tt{p(-1) = 2 - a + b }$

$\tt{\red{\implies{ b = a - 2}}}{\tt{\longrightarrow{\green{Equation - 1}}}}$

Similarly,

(x - 1) is a factor

So Let ,

x - 1 = 0

x = 1

Substitute this value in place of x in p(x)

$\tt{ p(1) = {(1)}^{4} + a{(1)}^{3} - 3{(1)}^{3} + 2(1) + b }$

$\tt{ p(1) = 1 + a(1) - 3(1) + 2 + b}$

$\tt{ p(1) = 1 + a - 3 + 2 + b }$

$\longrightarrow{\tt{\red{Substitute\; the\; value\; of \;b\; from\; equation\; 1 \;}}}$

$\tt{ p(1) = 1 + a - 3 + 2 + a - 2 }$

Here , +2 & -2 will get cancelled due to their opposite signs

$\tt{ p(1) = 1 + a - 3 + a }$

$\tt{ p(1) = -2 + 2a }$

$\Rightarrow{\tt{ -2 + 2a = 0 }}$

$\rightarrow{\tt{ 2a = 2 }}$

$\rightarrow{\tt{ a = \dfrac{2}{2}}}$

$\orange{\implies{\tt{ a = 1}}}$

Now,

Consider equation - 1

$\tt{ b = + a - 2}$

Here substitute the value of a in it

$\tt{ b = 1 - 2}$

$\orange{\implies{\tt{ b = - 1}}}$

$\rule{400}{4}$

Therefore ,

$\boxed{\blue{\tt{ Value \; of \; a \; = \; 1}}}$

$\boxed{\green{\tt{Value \; of \; b \; = \; - 1 }}}$

$\rule{400}{4}$