Question

use identity $( x + a ) ( x + b) = x² + ( a + b) x + ab$ to find the product of the following . $(1) ( x + 3 ) ( x + 7 ) . (2) ( 2a² + a) ( 2a² + 5 )$​

Answer 1

» 70,000 + 5,90,000 - 60,000

⇒ 6,00,000

C = 6,00,000

Gross t = Net S - Cost

⇒ Gross = 7,20,000 - 6,00,000

⇒ 1,20,000

Gross =1,20,000

Answer 2

Question :

Find the product of the following . [ by using identity (x+a)(x+b)=x² + ( a + b) x + ab ]

( x + 3 ) ( x + 7 )

( 2a² + a) ( 2a² + 5 )

Solution :

$\sf\fbox{We have to find the product}$

1) $( x + 3 ) ( x + 7 )$

By using identity $[ (x+a) (x+b)=x² + ( a + b) x + ab ]$

$\sf\implies\:(x+3)(x+7)$

$\sf=x^2+(7+3)x+7$

$\sf=x^2+10x+21=x$

Thus ,

$( x + 3 ) ( x + 7 ) = x ²+ 10 x +21$

2) $( 2a² + a) ( 2a² + 5 )$

Now , use identity $[ (x+a) (x+b)=x² + ( a + b) x + ab ]$

Then ,

$\sf\implies(2a^2+a)(2a^2+5)$

$\sf=(2a^2)^2+(a+5)(2a^2)$

$\sf=4a^4+2a^3+10a^2+5a=4a$

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More Algeraic Indentities:

1)$\sf{(a+b)^2=a^2+b^2+2ab}(a+b)$

2$)\sf{(a-b)^2=a^2+b^2-2ab}(a−b)$

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

4)$\sf{(a+b+c)^2={a}^{2}+{b}^{2}+{c}^{2}+2ab+2bc+2ca}4)(a+b+c)$

5)$\sf{(a+b)^3=a^3+b^3+3ab(a+b)}(a+b)$