Question

find the HCF and LCM f (x)=24(x^3+9x^2+20x) g(x)= 28(x^4+x^3-12x^2)​

Answer 1

Step-by-step explanation:

Given :-

f(x) = 24(x³+9x²+20x)

g(x) = 28(x⁴+x³-12x²)

To find :-

Find the HCF and the LCM of f(x) and g(x) ?

Solution :-

Given expressions are

f(x) = 24(x³+9x²+20x)

It can be written as

=> f(x) = 4×6×x(x²+9x+20)

=> f(x) = 4×6×x×(x²+5x+4x+20)

=> f(x) = 4×6×x×[(x(x+5)+4(x+5)]

=> f(x) = 4×6×x×(x+5)(x+4)

=> f(x) = 4×6×x×(x+5)×(x+4)

and

g(x) = 28(x⁴+x³-12x²)

=> g(x) = 4×7×x²×(x²+x-12)

=> g(x) = 4×7×x²×(x²+4x-3x-12)

=> g(x) = 4×7×x²×[x(x+4)-3(x+4)]

=> g(x) = 4×7×x²×(x+4)(x-3)

=> g(x) = 4×7×x×x×(x+4)(x-3)

=> g(x) = 4×7×x×x×(x+4)×(x-3)

We know that

f(x) = 4×6×x×(x+5)×(x+4)

g(x) = 4×7×x×x×(x+4)×(x-3)

HCF is the product of prime factors with least powers in the expressions

=> HCF of f(x) and g(x) = 4×x×(x+4)

=> HCF of f(x) and g(x) = 4x(x+4)

=> HCF of f(x) and g(x) =4x²+16x

LCM is the product of each prime factors with heighest powers

=> LCM of f(x) and g(x) = 4×6×7×x×x×(x-3)×(x+4)×(x+5)

=> LCM of f(x) and g(x) =168×x²×(x-3)(x+4)(x+5)

=> LCM of f(x) and g(x) = 168×x²×(x-3)×(x²+9x+20)

=> LCM of f(x) and g(x) = 168×x²×(x³+9x²+20x-3x²-27x-60)

=> LCM of f(x) and g(x) = 168×x²×(x³+6x²-7x-60)

=> LCM of f(x) and g(x) = 168(x⁵+6x⁴-7x³-60x²)

Answer:-

The HCF of the given expressions is 4x²+16x

The LCM of the given expressions is 168(x⁵+6x⁴-7x³-60x²)

Used formulae:-

→ LCM is the product of each prime factors with heighest powers

→ HCF is the product of prime factors with least powers in the expressions

→ a^m × a^n = a^(m+n)