Advertisements
Advertisements
Question
Express each of the following product as a monomials and verify the result in each case for x = 1:
(4x2) × (−3x) × \[\left( \frac{4}{5} x^3 \right)\]
Advertisements
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:
\[\left( 4 x^2 \right) \times \left( - 3x \right) \times \left( \frac{4}{5} x^3 \right)\]
\[ = \left\{ 4 \times \left( - 3 \right) \times \frac{4}{5} \right\} \times \left( x^2 \times x \times x^3 \right)\]
\[ = \left\{ 4 \times \left( - 3 \right) \times \frac{4}{5} \right\} \times \left( x^{2 + 1 + 3} \right)\]
\[ = - \frac{48}{5} x^6\]
\[\therefore\] \[\left( 4 x^2 \right) \times \left( - 3x \right) \times \left( \frac{4}{5} x^3 \right) = - \frac{48}{5} x^6\]
Substituting x = 1 in LHS, we get:
\[\text { LHS } = \left( 4 x^2 \right) \times \left( - 3x \right) \times \left( \frac{4}{5} x^3 \right)\]
\[ = \left( 4 \times 1^2 \right) \times \left( - 3 \times 1 \right) \times \left( \frac{4}{5} \times 1^3 \right)\]
\[ = 4 \times \left( - 3 \right) \times \frac{4}{5}\]
\[ = - \frac{48}{5}\]
Putting x = 1 in RHS, we get:
\[\text { RHS } = - \frac{48}{5} x^6 \]
\[ = - \frac{48}{5} \times 1^6 \]
\[ = - \frac{48}{5}\]
\[\because\] LHS = RHS for x = 1; therefore, the result is correct
Thus, the answer is \[- \frac{48}{5} x^6\].
APPEARS IN
RELATED QUESTIONS
Find each of the following product:
−3a2 × 4b4
Find each of the following product:
(7ab) × (−5ab2c) × (6abc2)
Find each of the following product: \[\left( \frac{4}{3}p q^2 \right) \times \left( - \frac{1}{4} p^2 r \right) \times \left( 16 p^2 q^2 r^2 \right)\]
Evaluate each of the following when x = 2, y = −1.
\[(2xy) \times \left( \frac{x^2 y}{4} \right) \times \left( x^2 \right) \times \left( y^2 \right)\]
Simplify:
(3x − 2)(2x − 3) + (5x − 3)(x + 1)
Show that: \[\left( \frac{4m}{3} - \frac{3n}{4} \right)^2 + 2mn = \frac{16 m^2}{9} + \frac{9 n^2}{16}\]
Show that: (4pq + 3q)2 − (4pq − 3q)2 = 48pq2
Show that: (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0
Multiply:
23xy2 × 4yz2
Solve the following equation.
5(x + 1) = 74
