Answer
Verified
395.4k+ views
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 \[GCD\left( {a,b} \right) \times LCM\left( {a,b} \right) = a \times b\].
Complete step by step answer:
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.
Complete step by step answer:
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.
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE