Find the prime factors of the following number:

$135$

If the HCF of polynomials $f\left(x\right)=\left(x-1\right)\left(x^2+3x+a\right)$ and $g\left(x\right)=\left(x+2\right)\left(x^2+2x+b\right)$ is $\left(x^2+x-2\right)$ then what are the values of $a$ and $b$ respectively?

The sum of all the factors of $156$ is: 
(A) $392$
(B) $235$
(C) $390$
(D) $379$

What is the number of Prime factor in ${15}^{18}$$\times$$4^{10}$$\times$$6^9$.

The total number of factors of 1156 is: 

(A) 9 

(B) 8 

(C) 10 

(D) 11 

The prime factorisation of 240 is:

(A) $2^3\times$$3^2\times 5$ 

(B) $2^4\times 3^2\times 5$

(C) $2^4\times 3\times 5$

(D) $2^5\times 3\times 5$

Find the HCF of 3, 7 and 15. 

(A) 7 

(B) 15 

(C) 3 

(D) 1