# Find each of the following products:

Question:

Find each of the following products:

(i) (x + 6)(x + 6)

(ii) (4x + 5y)(4x + 5y)

(iii) (7a + 9b)(7a + 9b)

(iv) $\left(\frac{2}{3} x+\frac{4}{5} y\right)\left(\frac{2}{3} x+\frac{4}{5} y\right)$

(v) $\left(x^{2}+7\right)\left(x^{2}+7\right)$

(vi) $\left(\frac{5}{6} a^{2}+2\right)\left(\frac{5}{6} a^{2}+2\right)$

Solution:

(i) We have:

$(x+6)(x+6)$

$=(x+6)^{2}$

$=x^{2}+6^{2}+2 \times x \times 6$       $\left[\right.$ using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=x^{2}+36+12 x$

(ii) We have:

$(4 x+5 y)(4 x+5 y)$

$=(4 x+5 y)^{2}$

$=(4 x)^{2}+(5 y)^{2}+2 \times 4 x \times 5 y$     $\left[\right.$ using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=16 x^{2}+25 y^{2}+40 x y$

(iii) We have:

$(7 a+9 b)(7 a+9 b)$

$=(7 a+9 b)^{2}$

$=(7 a)^{2}+(9 b)^{2}+2 \times 7 a \times 9 b$  $\left[\right.$ using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=49 a^{2}+81 b^{2}+126 a b$

(iv)  We have:

$\left(\frac{2}{3} x+\frac{4}{5} y\right)\left(\frac{2}{3} x+\frac{4}{5} y\right)$

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

$=\left(\frac{2}{3} x\right)^{2}+\left(\frac{4}{5} y\right)^{2}+2 \times \frac{2}{3} x \times \frac{4}{5} y$    $\left[\right.$ using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=\frac{4}{9} x^{2}+\frac{16}{25} y^{2}+\frac{16}{15} x y$

(v) $\left(x^{2}+7\right)\left(x^{2}+7\right)$

$=\left(x^{2}+7\right)^{2}$

$=\left(x^{2}\right)^{2}+7^{2}+2 \times x^{2} \times 7$      $\left[\right.$ using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=x^{4}+49+14 x^{2}$

(vi) We have:

$\left(\frac{5}{6} a^{2}+2\right)\left(\frac{5}{6} a^{2}+2\right)$

$=\left(\frac{5}{6} a^{2}+2\right)^{2}$

$=\left(\frac{5}{6} a^{2}\right)^{2}+(2)^{2}+2 \times \frac{5}{6} a^{2} \times 2$ $\left[\right.$ using $\left.(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$

$=\frac{25}{36} a^{4}+4+\frac{10}{3} a^{2}$