Question

Express each of the following product as a monomials and verify the result in each case for x = 1:
(x²)³ × (2x) × (−4x) × (5)

Answer 1

We have to find the product of the expression in order to express it as a monomial.

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​$a^m\times a^n=a^{m+n}\text{and}{\left(a^m\right)}^n=a^{mn}$.

We have:

$$\begin{gathered} {\left(x^2\right)}^3\times \left(2x\right)\times \left(-4x\right)\times 5 \\ =\left(x^6\right)\times \left(2x\right)\times \left(-4x\right)\times 5 \\ =\left\{2\times \left(-4\right)\times 5\right\}\times \left(x^6\times x\times x\right) \\ =\left\{2\times \left(-4\right)\times 5\right\}\times \left(x^{6+1+1}\right) \\ =-40x^8 \end{gathered}$$

$\therefore$ ${\left(x^2\right)}^3\times \left(2x\right)\times \left(-4x\right)\times 5=-40x^8$

Substituting x = 1 in LHS, we get:​

$$\begin{gathered} \text{LHS}={\left(x^2\right)}^3\times \left(2x\right)\times \left(-4x\right)\times 5 \\ ={\left(1^2\right)}^3\times \left(2\times 1\right)\times \left(-4\times 1\right)\times 5 \\ =1^6\times 2\times \left(-4\right)\times 5 \\ =1\times 2\times \left(-4\right)\times 5 \\ =-40 \end{gathered}$$

Putting x = 1 in RHS, we get:​

$$\begin{gathered} \text{RHS}=-40x^8 \\ =-40{\left(1\right)}^8 \\ =-40\times 1 \\ =-40 \end{gathered}$$

$\because$ LHS = RHS for x = 1; therefore, the result is correct

Thus, the answer is $-40x^8$.