Question

Solve $\left( {4x + 5} \right)\left( {4x - 1} \right)$ .

Answer 1

Hint: In this question, we are given two polynomials written in product with each other. So, we have to multiply these brackets with each other to solve it. We can do this using the distributive property.
Distributive property over addition is given by $a \times \left( {b + c} \right) = a \times b + a \times c$ .

Complete step-by-step answer:
Given equation $\left( {4x + 5} \right)\left( {4x - 1} \right)$ .
To solve the given equation by multiplying the brackets.
First, let us write the given equation $\left( {4x + 5} \right)\left( {4x - 1} \right)$ .
Now, to apply the distributive property we must have three terms but this equation has four right now.
So, consider $4x - 1$ as one term, and then using distributive property, we get, $4x(4x - 1) + 5(4x - 1)$ .
Now, again we have to apply the distributive property on both the terms $4x(4x - 1)$ and $5(4x - 1)$ separately, so, on applying, we get, $4x \times 4x - 4x \times 1 + 5 \times 4x - 5 \times 1$ .
On multiplying the terms, we get the following expression, $16{x^2} - 4x + 20x - 5$ .
Now, since, $ - 4x$ and $20x$ are like terms, and we know that $ - + = - $ , so on subtracting them, we get the following expression: $16{x^2} + 16x - 5$ .
Hence, on solving, $\left( {4x + 5} \right)\left( {4x - 1} \right)$ , we get, $16{x^2} + 16x - 5$ .

Note: Like terms are referred to as the terms with the same variable and with the same power. For example: $4{x^2}$ and $7{x^2}$ are like terms whereas, $4{x^2}$ and $4x$ or $4{x^2}$ and $4xy$ are unlike terms.
For this question, the distributive property can also be written as \[\left( {ax + b} \right)\left( {cx + d} \right) = ax\left( {cx + d} \right) + b\left( {cx + d} \right)\] , then again, we have to apply the distributive property for both the terms separately.
To solve an equation with multiple operations, we must use the BODMAS rule to get the correct answer. ‘BODMAS’ means B- Brackets, O- Of, D- Division, M- Multiplication, A- Addition, S- Subtraction.