Question

29
Use the identity (x + a) (x + b) = x2 + (a + b)x + ab to find the
following products.
a. (3.x + 2)(3x + 5)
b. (7x + 5) (7x - 1)
b. (4a2 + 1) (4a² + 3)
d. (abc - 2)(abc - 1)​

Answer 1

Answer :-

• According two question we have to find the product of the following numbers by using the given identity

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

• This identity comes after the simple multiplication of numbers.

1. (3x+2)(3x+5)

• x = 3x
• a = 2
• b = 5

$\bf{\ x^{2}+(a+b)x+ab}$

Using the given Identity

$\implies\sf{\ (3x)^{2}+(2+5)3x+(2)(5)}$

$\implies\sf{\ 9x^{2}+(7)3x+10}$

$\implies\sf{\ 9x^{2}+21x+10}$

2. (7x+5)(7x-1)

• x = 7x
• a = 5
• b = -1

$\bf{\ x^{2}+(a+b)x+ab}$

Using the given Identity

$\implies\sf{\ (7x)^{2}+(5-1)3x+(5)(-1)}$

$\implies\sf{\ 49x^{2}+(4)3x-5}$

$\implies\sf{\ 49x^{2}+12x-5}$

3. (4$\sf{a\ ^{2}}$+1)(4$\sf{a\ ^{2}}$+3)

• x = $\sf{4\ a^{2}}$
• a = 1
• b = 3

$\bf{\ x^{2}+(a+b)x+ab}$

Using the given Identity

$\implies\sf{\ (\ 4a^{2})^{2}+(1+3)\ 4a^{2}+(1)(3)}$

$\implies\sf{\ 16a^{4}+(4)\ 4a^{2}+3}$

$\implies\sf{\ 16a^{4}+\ 16a^{2}+3}$

4. (abc - 2)(abc - 1)

• x = abc
• a = -2
• b = -1

$\bf{\ x^{2}+(a+b)x+ab}$

Using the given Identity

$\implies\sf{\ (abc)^{2}+(-2-1)abc+(-2)(-1)}$

$\implies\sf{\ a^{2}\ b^{2}\ c^{2}+(-3)abc+2}$

$\implies\sf{\ a^{2}\ b^{2}\ c^{2}-3abc+2}$

$\bf{\underline{\red{Point\: to\: remember}}}$

• In multiplication if the numbers are same so we can write the whole square of that number .

Ex. = (2x)(2x) = $\sf{\ (2x)^{2}}$

Answer 2

Required to find :-

• Product of the given numbers

Mentioned condition :-

Use the Identity ;

• ( x + a ) ( x + b ) = x² + x ( a + b ) + ab

Solution :-

a.

Question :-

Find the product of ( 3x + 2 ) ( 3x + 5 )

Answer :-

( 3x + 2 ) ( 3x + 5 )

This is in the form of an identity ;

• ( x + a ) ( x + b ) = x² + x ( a + b ) + ab

So,

( 3x + 2 ) ( 3x + 5 ) =

=> ( 3x )² + 3x ( 2 + 5 ) + ( 2 ) ( 5 )

=> 9x² + 3x ( 7 ) + 10

=> 9x² + 21x + 10

Hence,

( 3x + 2 ) ( 3x + 5 ) = 9x² + 21x + 10

b .

Question :-

Find the product of ( 7x + 5 ) ( 7x - 1 )

Answer :-

( 7x + 5 ) ( 7x - 1 )

( 7x + 5 ) ( 7x + [ - 1 ] )

This is in the form of an identity ;

• ( x + a ) ( x + b ) = x² + x ( a + b ) + ab

So,

( 7x + 5 ) ( 7x + [ - 1 ] ) =

=> ( 7x )² + 7x ( 5 + [ - 1 ] ) + ( 5 ) ( - 1 )

=> 49x² + 7x ( 5 - 1 ) + ( - 5 )

=> 49x² + 7x ( 4 ) - 5

=> 49x² + 28x - 5

Hence,

( 7x + 5 ) ( 7x + [ - 1 ] ) = 49x² + 28x - 5

c .

Question :-

Find the product of ( 4a² + 1 ) ( 4a² + 3 )

Answer :-

( 4a² + 1 ) ( 4a² + 3 )

This is in the form of an identity ;

• ( x + a ) ( x + b ) = x² + x ( a + b ) + ab

( 4a² + 1 ) ( 4a² + 3 ) =

=> ( 4a² )² + 4a² ( 1 + 3 ) + ( 1 ) ( 3 )

=> 4a⁴ + 4a² ( 4 ) + 3

=> 4a⁴ + 16a² + 3

Hence,

( 4a² + 1 ) ( 4a² + 3 ) = 4a⁴ + 16a² + 3

d .

Question :-

Find the product of ( abc - 2 ) ( abc - 1 )

Answer :-

( abc - 2 ) ( abc - 1 )

( abc + [ - 2 ] ) ( abc + [ - 1 ] )

This is in the form of an identity ;

• ( x + a ) ( x + b ) = x² + x ( a + b ) + ab

So,

( abc + [ - 2 ] ) ( abc + [ - 1 ] ) =

=> ( abc )² + abc ( - 2 + [ - 1 ] ) + ( - 2 ) ( - 1 )

=> a²b²c² + abc ( - 2 - 1 ) + 2

=> a²b²c² + abc ( - 3 ) + 2

=> a²b²c² + ( - 3abc ) + 2

=> a²b²c² - 3abc + 2

Hence,

( abc - 2 ) ( abc - 1 ) = a²b²c² - 3abc + 2