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# The GCD of ${x^4} + 3{x^3} + 5{x^2} + 26x + 56$ and ${x^4} + 2{x^3} - 4{x^2} - x + 28$ is ${x^2} + 5x + 7$. Find their LCM.

Last updated date: 20th Jun 2024
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Hint: In this particular problem apply the formula that if a and b are the two numbers or polynomial equation then the relation between the GCD and LCM of the two numbers/equations is $GCD\left( {a,b} \right) \times LCM\left( {a,b} \right) = a \times b$.

As we know that LCM of two equations is the least common multiple of both the equations.
And the GCD of the equation is the greatest common divisor of both the equations.
So, let the given equations be f(x) and g(x).
So, f(x) = ${x^4} + 3{x^3} + 5{x^2} + 26x + 56$ (1)
And, g(x) = ${x^4} + 2{x^3} - 4{x^2} - x + 28$ (2)
So, now as we know the formula that if there are two polynomial equations then the product of their GCD and LCM is equal to the product of two equations.
So, now we are given that the GCD of f(x) and g(x) is ${x^2} + 5x + 7$.
Let, h(x) = ${x^2} + 5x + 7$ (3)
So, Let the LCM of the equation f(x) and g(x) be l(x)
So, according to the above stated formula $h\left( x \right) \times l\left( x \right) = f\left( x \right) \times g\left( x \right)$
Dividing both the sides of the above equation by h(x).
LCM = $l\left( x \right) = \dfrac{{f\left( x \right) \times g\left( x \right)}}{{h\left( x \right)}} = \dfrac{{\left( {{x^4} + 3{x^3} + 5{x^2} + 26x + 56} \right) \times \left( {{x^4} + 2{x^3} - 4{x^2} - x + 28} \right)}}{{\left( {{x^2} + 5x + 7} \right)}}$ (1)
Now let us divide f(x) by h(x) to simplify the above equation
${x^2} + 5x + 7\mathop{\left){\vphantom{1\begin{gathered} {x^4} + 3{x^3} + 5{x^2} + 26x + 56 \\ {x^4} + 5{x^3} + 7{x^2} \\ {\text{ }} - 2{x^3} - 2{x^2} + 26x + 56 \\ {\text{ }} - 2{x^3} - 10{x^2} - 14x \\ {\text{ }}8{x^2} + 40x + 56 \\ {\text{ }}8{x^2} + 40x + 56{\text{ }} \\ {\text{ 0 }} \\ \end{gathered} }}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{gathered} {x^4} + 3{x^3} + 5{x^2} + 26x + 56 \\ {x^4} + 5{x^3} + 7{x^2} \\ {\text{ }} - 2{x^3} - 2{x^2} + 26x + 56 \\ {\text{ }} - 2{x^3} - 10{x^2} - 14x \\ {\text{ }}8{x^2} + 40x + 56 \\ {\text{ }}8{x^2} + 40x + 56{\text{ }} \\ {\text{ 0 }} \\ \end{gathered} }}} \limits^{\displaystyle \,\,\, {{x^2} - 2x + 8}}$
So, the quotient when f(x) is divided by h(x) will be ${x^2} - 2x + 8$.
So, equation 1 becomes,
$\Rightarrow l\left( x \right) = \left( {{x^2} - 2x + 8} \right) \times \left( {{x^4} + 2{x^3} - 4{x^2} - x + 28} \right)$.
Hence the LCM of the ${x^4} + 3{x^3} + 5{x^2} + 26x + 56$ and ${x^4} + 2{x^3} - 4{x^2} - x + 28$ is $\left( {{x^2} - 2x + 8} \right) \times \left( {{x^4} + 2{x^3} - 4{x^2} - x + 28} \right)$.

Note:
Whenever we face such types of problems then first, we have to recall the definition of GCD and LCM of numbers (this will be the same for equations also). And after that we can apply the direct formula $GCD\left( {a,b} \right) \times LCM\left( {a,b} \right) = a \times b$ to find the GCD or LCM if one of them is already given. So, to find the LCM we had to divide the product of the numbers by their GCD and for that we can use the long division method. This will be the easiest and efficient way to find the solution of the problem.