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
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)
C.
(x-1)(x-2)(x-3)
Putting x =1 in polynomial
x3-6x2+11x-6,
f(1)=0
(1)3-6(1)2+11x1-6=0
1-6+11-6=0
12-12=0
0=0
Hence, x - l is a factor of polynomial x3-6x2+11x-6.
Now,
therefore(x-1)(x-5x+ 6)
=(x -1 )(x2- 3x - 2x + 6)
=(x -l)[x(x - 3)- 2(x - 3)]
=(x -l)(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