Use the identity\[(x + a)(x + b) = {x^2} + (a + b)x + ab\]to find the following products.
(i) \[(x + 3)(x + 7)\]
(ii) $(4x + 5)(4x + 1)$
(iii) $(4x - 5)(4x - 1)$
(iv) $(4x + 5)(4x - 1)$
(v) $(2x + 5y)(2x + 3y)$
(vi) \[(2{a^2} + 9)(2{a^2} + 5)\]
(vii) \[(xyz - 4)(xyz - 2)\]

Hint:Substitute the values in the identity \[(x + a)(x + b) = {x^2} + (a + b)x + ab\] to get a polynomial. Add\[a\]and$b$ and simplify the polynomial to get the final answer.

Complete step-by-step solution
We are given 7 pairs of binomials.
We need to find the product of the binomials using the identity \[(x + a)(x + b) = {x^2} + (a + b)x + ab....(1)\] in each of these pairs.
(i) Consider the product \[(x + 3)(x + 7)\]
Comparing with (1), we can see that here $x = x$, $a = 3$ and $b = 7$
Thus, we have \[(x + 3)(x + 7) = {x^2} + (3 + 7)x + 3 \times 7 = {x^2} + 10x + 21\]
Hence the product is\[{x^2} + 10x + 21\].

 (ii) Consider the product $(4x + 5)(4x + 1)$
Comparing with (1), we can see that here $x = 4x$, $a = 5$ and $b = 1$
Thus, we have \[(4x + 5)(4x + 1) = {(4x)^2} + (5 + 1)4x + 5 \times 1 = 16{x^2} + 6 \times 4x + 5 = 16{x^2} + 24x + 5\]
Hence the product is\[16{x^2} + 24x + 5\].

(iii) Consider the product $(4x - 5)(4x - 1)$
We can rewrite this product as$(4x + ( - 5))(4x + ( - 1))$
Comparing with (1), we can see that here $x = 4x$, $a = - 5$ and $b = - 1$
Thus, we have \[(4x + ( - 5))(4x + ( - 1)) = {(4x)^2} + (( - 5) + ( - 1))4x + ( - 5 \times - 1) = 16{x^2} - 6 \times 4x + 5 = 16{x^2} - 24x + 5\]
Hence the product is \[16{x^2} + 24x + 5\].

(iv) Consider the product $(4x + 5)(4x - 1)$
Comparing with (1), we can see that here $x = 4x$, $a = 5$ and $b = - 1$
Thus, we have \[(4x + 5)(4x - 1) = {(4x)^2} + (5 + ( - 1))4x + 5 \times ( - 1) = 16{x^2} + 4 \times 4x - 5 = 16{x^2} + 16x - 5\]
Hence the product is \[16{x^2} + 16x - 5\].

(v) Consider the product $(2x + 5y)(2x + 3y)$
Comparing with (1), we can see that here $x = 2x$, $a = 5y$ and $b = 3y$
Thus, we have \[(2x + 5y)(2x + 3y) = {(2x)^2} + (5y + 3y)2x + 5y \times 3y = 4{x^2} + 8y \times 2x + 15{y^2} = 4{x^2} + 16xy + 15{y^2}\]
Hence the product is\[4{x^2} + 16xy + 15{y^2}\].

(vi) Consider the product \[(2{a^2} + 9)(2{a^2} + 5)\]
Comparing with (1), we can see that here $x = 2{a^2}$, $a = 9$ and $b = 5$
Thus, we have \[(2{a^2} + 9)(2{a^2} + 5) = {(2{a^2})^2} + (9 + 5)2{a^2} + 9 \times 5 = 4{a^4} + 14 \times 2{a^2} + 45 = 4{a^4} + 28{a^2} + 45\]
Hence the product is\[4{a^4} + 28{a^2} + 45\].

(vii) Consider the product \[(xyz - 4)(xyz - 2)\]
Comparing with (1), we can see that here $x = xyz$, $a = - 4$ and $b = - 2$
Thus, we have \[(xyz - 4)(xyz - 2) = {(xyz)^2} + (( - 4) + ( - 2))xyz + ( - 4 \times - 2) = {x^2}{y^2}{z^2} + - 6xyz + 8\]
Hence the product is \[{x^2}{y^2}{z^2} + - 6xyz + 8\].

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 $(4x - 3{x^3})$.