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
If HCF of (p2- 8p + 12) and (p2 + 4p -12) is (p-a), then the value of a is
-6
2
-2
6
B.
2
p-a=0
p=a
On putting the value of 'p' in both polynomials, we get
a2-8a+12 =0
a2-6a-2a+12 =0
a(a-6)-2(a-6)=0
(a-2)(a-6)=0
a=2,6
Again, a2+ 4a-12=0
a2+ 6a-2a-12 =0
a(a + 6)-2(a+ 6) =0
(a-2)(a+6)=0
a=2,-6
Hence, a=2
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