The polynomial ax3 + bx2 + x - 6 has (x+ 2) as a factor and leaves remainder 4 when divided by (x-2) Find the value of a and b.
a=2,b=0
a=1,b=1
a=0,b=2
a=7,b=5
C.
a=0,b=2
Given, (x + 2) is a factor of polynomial ax3 + bx2 + x- 6
Then, x + 2 = 0
x=-2
f(x)=0, f(-2)=0
.......(1)
Polynomial ax3 + bx2+ x- 6 leaves remainder 4 when divided by(x - 2)
Then, x-2=0
x=2
f (x) =0, f (2)=0
.......(2)
From equation 1 and 2 we get a= 0 and b = 2.
A non-terminating but recurring decimal is
an integer
a natural number
a rational number
an irrational number
The factors of polynomial x3- 6x2 + 11x - 6 are
(x+1)(x+2)(x-3)
(x-1)(x+2)(x-3)
(x-1)(x-2)(x-3)
(x-1)(x-2)(x+3)
If a, a + 2, a + 4 are prime numbers, then the number of values of a is
One
two
three
more than three