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:

\[\left( x^2 \right)^3 \times \left( 2x \right) \times \left( - 4x \right) \times 5\]

\[ = \left( x^6 \right) \times \left( 2x \right) \times \left( - 4x \right) \times 5\]

\[ = \left\{ 2 \times \left( - 4 \right) \times 5 \right\} \times \left( x^6 \times x \times x \right)\]

\[ = \left\{ 2 \times \left( - 4 \right) \times 5 \right\} \times \left( x^{6 + 1 + 1} \right)\]

\[ = - 40 x^8 \]

\[\therefore\] \[\left( x^2 \right)^3 \times \left( 2x \right) \times \left( - 4x \right) \times 5 = - 40 x^8\]

Substituting x = 1 in LHS, we get:​

\[\text { LHS } { = \left( x^2 \right)^3 \times \left( 2x \right) \times \left( - 4x \right) \times 5\]

\[ = \left( 1^2 \right)^3 \times \left( 2 \times 1 \right) \times \left( - 4 \times 1 \right) \times 5\]

\[ = 1^6 \times 2 \times \left( - 4 \right) \times 5\]

\[ = 1 \times 2 \times \left( - 4 \right) \times 5\]

\[ = - 40\]

Putting x = 1 in RHS, we get:​

\[\text { RHS } = - 40 x^8 \]

\[ = - 40 \left( 1 \right)^8 \]

\[ = - 40 \times 1\]

\[ = - 40\]

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

Thus, the answer is \[- 40 x^8\].