if [x+2] is a factor of p[x]=ax^3+bx^2+x-6 and p[x], when divided by [x-2] leaves a remainder of 4, prove that a=0 and b=2
Answers
SOLUTION
Case 1
p(x)= ax^3+bx^2 +x -6
x+2 is factor of p(x)
=) x+2=0
=) x= -2
=) a(-2)^3 +b(-2)^2 +(-2) -6
=) -8a + 4b -2-6= 0
=) -8a+4b-8= 0
=) -8a+4b= 8
=) -2a+ b= 2
=) b= 2+2a...........(1)
Case 2
Since it's given that when p(x) is divided by x-2 it leaves as a remainder of 4,
Since it's given that when p(x) is divided by x-2 it leaves as a remainder of 4,so by remainder theorem we get
p(2)= 4
p(2)= 4=) 8a+4b+2-6= 4
=) 8a+4b= 8
=) 2a+b= 2
=) 2a +2+ 2a= 2 { putting eq. 1 place of b}
=) 4a= 2-2
=) 4a= 0
=a= 0
so,
putting the value of a in eq^n (1)
=) b= 2+2a
=)b= 2+ 2(0)
=) b= 2+0
=) b= 2
Hence,
a= 0
b= 2
HOPE it helps ✔️
Answer:
Step-by-step explanation:
Given :-
(x+2) is a factor of p(x)=ax^3+bx^2+x-6 and p(x) when divided by(x-2) leaves the remainder 4.
To find :-
Prove that a=0 and b=2
Solution :-
Given Polynomial is ax³+bx²+x-6
Let P(x) = ax³+bx²+x-6
Given factor of P(x) = (x+2)
we know that ,by Factor Theorem
If (x-a) is a factor of P(x) then P(a) = 0
Now,
If (x+2) is a factor of P(x) then P(-2) = 0
=> a(-2)³+b(-2)²+(-2)-6 = 0
=> a(-8)+b(4)-2-6 = 0
=> -8a+4b-8 = 0
=> 4(-2a+b-2) = 0
=> -2a+b-2 = 0/4
=> -2a+b-2= 0
=> b = 2a+2 -----------(1)
and
Given that
P(x) is divided by(x-2) leaves the remainder 4.
By Remainder Theorem
If P(x) is divided by x-a then the remainder is P(a).
Now,
P(x) is divided by(x-2) then the remainder = P(2).
According to the given problem.
P(2) = 4
=> a(2)³+b(2)²+2-6 = 4
=> a(8)+b(4)-4 = 4
=> 8a+4b-4 = 4
=> 8a+4b = 4+4
=> 8a+4b = 8
=> 4(2a+b) = 8
=> 2a+b = 8/4
=>2a+b = 2
=> 2a+2a+2 = 2
=> 4a +2 = 2
=> 4a = 2-2
=> 4a = 0
=> a = 0/4
=> a = 0
On substituting the value of a in (1) then
=> b = 2a+2
=> b = 2(0)+2
=> b = 0+2
=> b = 2
Therefore, a = 0 and b = 2