Express each of the following product as a monomials and verify the result in each case for x = 1:
Question:

Express each of the following product as a monomials and verify the result in each case for x = 1:

$\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$

Solution:

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 law of indices, i.e., $a^{m} \times a^{n}=a^{m+n}$.

We have:

$\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$

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

$=\left\{4 \times(-3) \times \frac{4}{5}\right\} \times\left(x^{2+1+3}\right)$

$=-\frac{48}{5} x^{6}$

$\therefore\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)=-\frac{48}{5} x^{6}$

Substituting x = 1 in LHS, we get:

$\mathrm{LHS}=\left(4 x^{2}\right) \times(-3 x) \times\left(\frac{4}{5} x^{3}\right)$

$=\left(4 \times 1^{2}\right) \times(-3 \times 1) \times\left(\frac{4}{5} \times 1^{3}\right)$

$=4 \times(-3) \times \frac{4}{5}$

$=-\frac{48}{5}$

Putting x = 1 in RHS, we get:

RHS $=-\frac{48}{5} x^{6}$

$=-\frac{48}{5} \times 1^{6}$

$=-\frac{48}{5}$

$\because L H S=\mathrm{RHS}$ for $x=1$; therefore, the result is correct

Thus, the answer is $-\frac{48}{5} x^{6}$.

 

 

 

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