Question

Find the product
(i) \[(5 - 2x)(3 + x)\]
(ii) $(x + 7y)(7x - y)$
(iii) $({a^2} + b)(a + {b^2})$
(iv) $({p^2} - {q^2})(2p + q)$

Answer 1

Hint:Multiply each term of the first bracket with each term of the second bracket to get a polynomial of 4 terms. Combine the like terms and simplify the polynomial to get the final answer.

Complete step-by -step solution:
We are given 4 pairs of binomials.
We need to find the product of the binomials in each of these pairs.
(i) Given binomials \[(5 - 2x)\] and \[(3 + x)\]
Consider their product\[(5 - 2x)(3 + x)\]
We will first multiply each term of the binomial\[(5 - 2x)\]with each term of the binomial\[(3 + x)\].
Thus, we have \[(5 - 2x)(3 + x) = 5(3 + x) - 2x(3 + x)\]
Here we will multiply the term outside a bracket with each of the terms inside that bracket to obtain a polynomial with 4 terms.
This gives us\[(5 - 2x)(3 + x) = (5 \times 3 + 5 \times x) - (2x \times 3 + 2x \times x) = (15 + 5x) - (6x + 2{x^2})\]
Now we will combine the like terms (the underlined part) to simplify the equation and get the final answer.
\[(5 - 2x)(3 + x) = 15\underline { + 5x - 6x} - 2{x^2} = 15 - x - 2{x^2}\]
Hence the product of\[(5 - 2x)\] and \[(3 + x)\]is \[15 - x - 2{x^2}\].

We will be repeating the above method for the next 3 pairs of binomials.
(ii) Consider the product$(x + 7y)(7x - y)$
$
  (x + 7y)(7x - y) = x(7x - y) + 7y(7x - y) \\
   = (x \times 7x - x \times y) + (7y \times 7x - 7y \times y) \\
   = (7{x^2} - xy) + (49xy - 7{y^2}) \\
   = 7{x^2}\underline { - xy + 49xy} - 7{y^2} \\
   = 7{x^2} + 48xy - 7{y^2} \\
 $
Hence the product is $7{x^2} + 48xy - 7{y^2}$.
(iii) Consider the product $({a^2} + b)(a + {b^2})$
$
  ({a^2} + b)(a + {b^2}) = {a^2}(a + {b^2}) + b(a + {b^2}) \\
   = ({a^2} \times a + {a^2} \times {b^2}) + (b \times a + b \times {b^2}) \\
   = ({a^3} + {a^2}{b^2}) + (ab + {b^3}) \\
   = {a^3} + {a^2}{b^2} + ab + {b^3} \\
 $
Hence the product is${a^3} + {a^2}{b^2} + ab + {b^3}$.
(iv) Consider the product $({p^2} - {q^2})(2p + q)$
$
  ({p^2} - {q^2})(2p + q) = {p^2}(2p + q) - {q^2}(2p + q) \\
   = ({p^2} \times 2p + {p^2} \times q) - ({q^2} \times 2p + {q^2} \times q) \\
   = (2{p^3} + {p^2}q) - (2p{q^2} + {q^3}) \\
   = 2{p^3} + {p^2}q - 2p{q^2} - {q^3} \\
 $
Hence the product is $2{p^3} + {p^2}q - 2p{q^2} - {q^3}$.

Note: If there is a minus sign outside the bracket, then while opening the bracket, the signs of the terms inside the bracket change.
For example, \[(25 + 6x) - (16x + 20{x^2}) = 25 + 6x - 16x - 20{x^2}\]