Find the value of a and b so that (x+1) (x-1)are factors of x4+ ax3- 3x2 +2x+b.
Answers
Step-by-step explanation:
put x=1
1+a-3+2+b=0
- a+b=0
put x=-1
1-a-3-2+b=0
b-a=4
- b=4+a.
a+4+a=0
2a=-4
a=-2
b=-a
b=2
Answer:
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}
Step-by-step explanation: