Answer
424.8k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)