Question

Find the LCM of the following: $10(9x^2+6xy+y^2),12(3x^2-5xy-2y^2),14(6x^4+2x^3)$

• A. $420x^3(3x+y{)}^2(x-2y\left)\right(3x+1)$
• B. $420x^3(3-y{)}^2(x-2y\left)\right(3x+1)$
• C. $420x^3(3x+y{)}^2(x+2y\left)\right(3x+1)$
• D. None of these

Answer 1

The correct option is A $420x^3(3x+y{)}^2(x-2y\left)\right(3x+1)$

We know that the least common multiple (LCM) is the smallest number or expression that is a common multiple of two or more numbers or algebraic terms.

We first factorize the given polynomials $10(9x^2+6xy+y^2),12(3x^2-5xy-2y^2)$ and $14(6x^4+2x^3)$ as shown below:

$10(9x^2+6xy+y^2)=10\left[\right(3x{)}^2+(2\times 3x\times y)+y^2\left)\right]=10\left[\right(3x+y{)}^2\left]\right(\because (a+b{)}^2=a^2+b^2+2ab)=2\times 5\times (3x+y)\times (3x+y)$

$12(3x^2-5xy-2y^2)=12(3x^2-6xy+xy-2y^2)=12\left[3x\right(x-2y)+y(x-2y\left)\right]=12(3x+y)(x-2y)=2\times 2\times 3\times (3x+y)\times (x-2y)$

$14(6x^4+2x^3)=14\times 2x^3(3x+1)=28x^3(3x+1)=2\times 2\times 7\times x\times x\times x\times (3x+1)$

Now multiply all the factors, using each common factor only once, therefore, the LCM is:

$2\times 2\times 3\times 5\times 7\times x\times x\times x\times (3x+y)\times (3x+y)\times (x-2y)\times (3x+1)=420x^3(3x+y{)}^2(x-2y\left)\right(3x+1)$

Hence, the LCM of $10(9x^2+6xy+y^2),12(3x^2-5xy-2y^2)$ and $14(6x^4+2x^3)$ is $420x^3(3x+y{)}^2(x-2y\left)\right(3x+1)$.