Find the product of the following binomials:
(i) $ (2 x+y)(2 x+y) $
(ii) $ (a+2 b)(a-2 b) $
(iii) $ \left(a^{2}+b c\right)\left(a^{2}-b c\right) $
(iv) $ \left(\frac{4 x}{5}-\frac{3 y}{4}\right)\left(\frac{4 x}{5}+\frac{3 y}{4}\right) $
(v) $ \left(2 x+\frac{3}{y}\right)\left(2 x-\frac{3}{y}\right) $
(vi) $ \left(2 a^{3}+b^{3}\right)\left(2 a^{3}-b^{3}\right) $
(vii) $ \left(x^{4}+\frac{2}{x^{2}}\right)\left(x^{4}-\frac{2}{x^{2}}\right) $
(viii) $ \left(x^{3}+\frac{1}{x^{3}}\right)\left(x^{3}-\frac{1}{x^{3}}\right) $.

To do:

We have to find the product of the given binomials.

Solution:

We know that,

$(a + b)^2 = a^2 + 2ab + b^2$

$(a - b)^2 = a^2 - 2ab + b^2$

Therefore,

(i) $(2 x+y)(2 x+y) = (2 x+y)^2$

$=(2x)^2+2(2x)(y)+(y)^2$

$= 4x^2 + 4xy + y^2$

(ii) $(a+2 b)(a-2 b) =(a)^2-(2b)^2$

$=a^2-4b^2$

(iii) $(a^2+bc)(a^2-bc) =( {a^2})^2-(bc)^2$

$=a^4-b^2c^2$

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

$=\frac{16x^2}{25}-\frac{9y^2}{16}$

(v) $(2x+\frac{3}{y})(2x-\frac{3}{y}) = (2x)^2-(\frac{3}{y})^2$

$=4x^2-\frac{9}{y^2}$

(vi)  $(2 a^{3}+b^{3})(2 a^{3}-b^{3}) = (2a^3)^2-(b^3)^2$

$=4a^6-b^6$

(vii) $(x^{4}+\frac{2}{x^2})({x}^4-\frac{2}{x^2}) = ({x}^4)^2-(\frac{2}{x^2})^2$

$=x^8-\frac{4}{x^4}$

(viii) $({x}^3+\frac{1}{x^3})({x}^3-\frac{1}{x^3}) = (x^3)^2-(\frac{1}{x^3})^2$

$=x^6-\frac{1}{x^6}$

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

22 Views

Kickstart Your Career

Get certified by completing the course

Get Started