Question

(a) Find the square of (4x2 – 7y2 ) using an appropriate identity.

Answer 1

Answer:

a2-b2=(a+b)(a-b)

a=4x, b=7y

(4x+7y)(4x-7y)

16x2-28xy+28xy- 49y2

16x2-49y2

Answer 2

Answer:

-33x⁴

Step-by-step explanation:

Given:

Polynomial : (4x² - 7y²)

To Find:

• Square of the polynomial.

Identity used:

• (a - b)² = a²

Procedure :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: {( {4x}^{2} - {7x}^{2} )}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: ( {4x}^{2} - {7x}^{2} )( {4x}^{2} - {7x}^{2} )$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: {(4 {x}^{2} )}^{2} - { {(7x}^{2} )}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: {16x}^{4} - {49x}^{4}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: - {33x}^{4}$

proceed towards all identities .

${(a + b)}^{2} = {a}^{2} + 2.a.b + {b}^{2}(a+b) 2=a 2+2.a.b+b 2$

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

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

${(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2.a.b + 2.b.c + 2.c.a(a+b+c) 2=a 2+b2+c 2 +2.a.b+2.b.c+2.c.a$

${(a + b)}^{3} = {a}^{3} + {b}^{3} + 3.a.b(a + b)(a+b) 3 =a 3+b 3+3.a.b(a+b)$

${(a - b)}^{3} = {a}^{3} - {b}^{3 } - 3.a.b (a - b)(a-b) 3=a 3 -b 3 -3.a.b(a-b)$

${a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2} )a 3+b3 =(a+b)(a 2-ab+b 2 )$

${a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} )a3-b 3=(a-b)(a 2 +ab+b2)$

${a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)a 3+b 3 +c3 =(a+b+c)(a2+b2+c 2-ab-bc-ca)$