Question

find the hcf and lemon of 30,72 and 423 by prime factorisation method​

Answer 1

p(x)=(x+1)(x

2

+ax+4) ........ (i)

and q(x)=(x+4)(x

2

+bx+2) ........ (ii)

h(x)=x

2

+5x+4 is HCF of p(x) and q(x)

h(x)=x

2

+5x+4

=x

2

+4x+x+4

=x(x+4)+(x+4)

=(x+4)(x+1)

∴x=4,−1 are the roots of p(x),q(x) and h(x)

∴p(x)=(x+1)(x

2

+ax+4)=(x+1)(x+4)f(x), for some f(x)

⟹(x

2

+ax+4)=(x+4)f(x)

Take x=−4

⟹((−4)

2

+a(−4)+4)=0

⟹16−4a+4=0

⟹a=5

Similarly, q(x)=(x+1)(x+4)g(x), for some factor g(x) of q(x)

⟹(x

2

+bx+2)=(x+1)g(x)

Take x=−1

⟹(−1)

2

+b(−1)+2=0

⟹1−b+2=0

⟹b=3

Hence, a=5 and b=3