Solve the following $\left( 4x+5y \right)\left( 4x-5y \right)$.
Answer
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Hint: To solve the question we need to know the concept of algebraic multiplication. The property of distribution is applied here to expand the two functions. Further in the question we will add the term having the same function like ${{x}^{2}},xy,{{y}^{2}}$. After multiplying the two functions, new functions originate.
Complete step by step solution:
The question asks us to multiply $\left( 4x+5y \right)\left( 4x-5y \right)$, which are functions in $x$ and $y$. To multiply both the functions we would distribute the terms. Let us assume two functions $(a+b)$ and $\left( p+q \right)$ , when both the function is multiplied to each other we get:
$\Rightarrow \left( a+b \right)\left( p+q \right)$
On expanding above function, we get:
$\Rightarrow a\left( p+q \right)+b\left( p+q \right)$
$\Rightarrow ap+aq+bp+bq$
This is what we get after expansion, applying the same to the question we have
$\Rightarrow \left( 4x+5y \right)\left( 4x-5y \right)$
On expanding the problem given to us, we get:
$\Rightarrow 4x\left( 4x-5y \right)+5y\left( 4x-5y \right)$
On multiplying the terms we get
$\Rightarrow \left( 16{{x}^{2}}-20xy \right)+\left( 20xy-25{{y}^{2}} \right)$
On removing the bracket we see that the term $20xy$ cancel out, as this term with both positive and negative sign is present, so the result comes out to be:
$\Rightarrow 16{{x}^{2}}-25{{y}^{2}}$
$\therefore $ The value of $\left( 4x+5y \right)\left( 4x-5y \right)$ is $16{{x}^{2}}-25{{y}^{2}}$.
Note: There is another approach to solve this question, which is shorter than the above one. When two functions are in form $\left( a+b \right)$ and $\left( a-b \right)$ then there is no need to expand the function to find the value of their product. We just need to apply a formula which is $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$ . The formula comes out to be this because in the multiplication of functions like this the middle term turns to $0$ that cancel out each other.
So applying this formula to the question we get:
$\Rightarrow \left( 4x+5y \right)\left( 4x-5y \right)=16{{x}^{2}}-25{{y}^{2}}$
We get the same answer on applying the formula, so these types of function could be solved with the above formula.
Complete step by step solution:
The question asks us to multiply $\left( 4x+5y \right)\left( 4x-5y \right)$, which are functions in $x$ and $y$. To multiply both the functions we would distribute the terms. Let us assume two functions $(a+b)$ and $\left( p+q \right)$ , when both the function is multiplied to each other we get:
$\Rightarrow \left( a+b \right)\left( p+q \right)$
On expanding above function, we get:
$\Rightarrow a\left( p+q \right)+b\left( p+q \right)$
$\Rightarrow ap+aq+bp+bq$
This is what we get after expansion, applying the same to the question we have
$\Rightarrow \left( 4x+5y \right)\left( 4x-5y \right)$
On expanding the problem given to us, we get:
$\Rightarrow 4x\left( 4x-5y \right)+5y\left( 4x-5y \right)$
On multiplying the terms we get
$\Rightarrow \left( 16{{x}^{2}}-20xy \right)+\left( 20xy-25{{y}^{2}} \right)$
On removing the bracket we see that the term $20xy$ cancel out, as this term with both positive and negative sign is present, so the result comes out to be:
$\Rightarrow 16{{x}^{2}}-25{{y}^{2}}$
$\therefore $ The value of $\left( 4x+5y \right)\left( 4x-5y \right)$ is $16{{x}^{2}}-25{{y}^{2}}$.
Note: There is another approach to solve this question, which is shorter than the above one. When two functions are in form $\left( a+b \right)$ and $\left( a-b \right)$ then there is no need to expand the function to find the value of their product. We just need to apply a formula which is $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$ . The formula comes out to be this because in the multiplication of functions like this the middle term turns to $0$ that cancel out each other.
So applying this formula to the question we get:
$\Rightarrow \left( 4x+5y \right)\left( 4x-5y \right)=16{{x}^{2}}-25{{y}^{2}}$
We get the same answer on applying the formula, so these types of function could be solved with the above formula.
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