Question

Multiply the binomials
(i) $(2x + 5)$ and $(4x - 3)$
(ii) $(y - 8)$and $(3y - 4)$
(iii) $(2.5l - 0.5m)$ and $(2.5l + 0.5m)$
(iv) $(a + 3b)$ and $(x + 5)$
(v) $(2pq + 3{q^2})$ and $(3pq - 2{q^2})$
(vi) $(\dfrac{3}{4}{a^2} + 3{b^2})$ and $({a^2} - \dfrac{2}{3}{b^2})$

Answer 1

Hint:Multiply each term of the first binomial with each term of the second binomial 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 eight pairs of binomials.
We need to find the product of binomials in each of these pairs.
(i) Given binomials $(2x + 5)$ and $(4x - 3)$
Consider their product$(2x + 5)(4x - 3)$
We will first multiply each term of the binomial$(2x + 5)$with each term of the binomial$(4x - 3)$.
Thus, we have$(2x + 5)(4x - 3) = 2x(4x - 3) + 5(4x - 3)$
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$(2x + 5)(4x - 3) = (2x \times 4x - 2x \times 3) + (5 \times 4x - 5 \times 3) = (8{x^2} - 6x) + (20x - 15)$
Now we will combine the like terms (the underlined part) to simplify the equation and get the final answer.
\[(2x + 5)(4x - 3) = 8{x^2}\underline { - 6x + 20x} - 15 = 8{x^2} + 14x - 15\]
Hence the product of$(2x + 5)$ and $(4x - 3)$is $8{x^2} + 14x - 15$.

We will be repeating the above method for the next 5 pairs of binomials.
(ii) $(y - 8)$and $(3y - 4)$
\[
  (y - 8)(3y - 4) = y(3y - 4) - 8(3y - 4) \\
   = (y \times 3y - y \times 4) - (8 \times 3y - 8 \times 4) \\
   = (3{y^2} - 4y) - (24y - 32) \\
   = 3{y^2}\underline { - 4y - 24y} + 32 \\
   = 3{y^2} - 28y + 32 \\
 \]
Hence the product of$(y - 8)$and $(3y - 4)$ is \[3{y^2} - 28y + 32\].
(iii) $(2.5l - 0.5m)$ and $(2.5l + 0.5m)$
$
  (2.5l - 0.5m)(2.5l + 0.5m) = 2.5l(2.5l + 0.5m) - 0.5m(2.5l + 0.5m) \\
   = (2.5l \times 2.5l + 2.5l \times 0.5m) - (0.5m \times 2.5l + 0.5m \times 0.5m) \\
   = (6.25{l^2} + 1.25lm) - (1.25lm + 0.25{m^2}) \\
   = 6.25{l^2}\underline { + 1.25lm - 1.25lm} - 0.25{m^2} \\
   = 6.25{l^2} - 0.25{m^2} \\
 $
Hence the product of $(2.5l - 0.5m)$ and $(2.5l + 0.5m)$ is $6.25{l^2} - 0.25{m^2}$.
(iv) $(a + 3b)$ and $(x + 5)$
$
  (a + 3b)(x + 5) = a(x + 5) + 3b(x + 5) \\
   = (a \times x + a \times 5) + (3b \times x + 3b \times 5) \\
   = (ax + 5a) + (3bx + 15b) \\
   = ax + 5a + 3bx + 15b \\
 $
Hence the product of $(a + 3b)$ and $(x + 5)$ is $ax + 5a + 3bx + 15b$.
(v) $(2pq + 3{q^2})$ and $(3pq - 2{q^2})$
$
  (2pq + 3{q^2})(3pq - 2{q^2}) = 2pq(3pq - 2{q^2}) + 3{q^2}(3pq - 2{q^2}) \\
   = (2pq \times 3pq - 2pq \times 2{q^2}) + (3{q^2} \times 3pq - 3{q^2} \times 2{q^2}) \\
   = (6{p^2}{q^2} - 4p{q^3}) + (9p{q^3} - 6{q^4}) \\
   = 6{p^2}{q^2}\underline { - 4p{q^3} + 9p{q^3}} - 6{q^4} \\
   = 6{p^2}{q^2} + 5p{q^3} - 6{q^4} \\
 $
Hence the product of $(2pq + 3{q^2})$ and $(3pq - 2{q^2})$ is $6{p^2}{q^2} + 5p{q^3} - 6{q^4}$.
(vi) $(\dfrac{3}{4}{a^2} + 3{b^2})$ and $({a^2} - \dfrac{2}{3}{b^2})$
$
  (\dfrac{3}{4}{a^2} + 3{b^2})({a^2} - \dfrac{2}{3}{b^2}) = \dfrac{3}{4}{a^2}({a^2} - \dfrac{2}{3}{b^2}) + 3{b^2}({a^2} - \dfrac{2}{3}{b^2}) \\
   = (\dfrac{3}{4}{a^2} \times {a^2} - \dfrac{3}{4}{a^2} \times \dfrac{2}{3}{b^2}) + (3{b^2} \times {a^2} - 3{b^2} \times \dfrac{2}{3}{b^2}) \\
   = (\dfrac{3}{4}{a^4} - \dfrac{1}{2}{a^2}{b^2}) + (3{a^2}{b^2} - 2{b^4}) \\
   = \dfrac{3}{4}{a^4}\underline { - \dfrac{1}{2}{a^2}{b^2} + 3{a^2}{b^2}} - 2{b^4} \\
   = \dfrac{3}{4}{a^4} + \dfrac{5}{2}{a^2}{b^2} - 2{b^4} \\
 $
Hence the product of $(\dfrac{3}{4}{a^2} + 3{b^2})$ and $({a^2} - \dfrac{2}{3}{b^2})$ is $\dfrac{3}{4}{a^4} + \dfrac{5}{2}{a^2}{b^2} - 2{b^4}$.

Note: A binomial is a polynomial with only two terms. It is an algebraic expression consisting of two monomials. That is, the two monomials in a binomial are separated by the operations of addition or subtraction.
Examples of binomials are $(2{x^2} + 5x)$ and $(4x - 3{x^3})$