Question

find the product by using a suitable identity.
(x + 5) (x + 4)
(a + 3) (a + 6)
(x-9)(x+7)​

Answer 1

Answer:

use the identity (x+a)(x+b)=x^2-xb-ax-ab.... u'll arrive with the answer

hope this helps u....

Answer 2

Answer:

Using suitable algebraic identities, find the product

Using suitable algebraic identities, find the product(A) (x+5)(x−5)

Using suitable algebraic identities, find the product(A) (x+5)(x−5)(B) (x−5)(x+5)

Using suitable algebraic identities, find the product(A) (x+5)(x−5)(B) (x−5)(x+5)(C) (3x+5)(3x+5)

Using suitable algebraic identities, find the product(A) (x+5)(x−5)(B) (x−5)(x+5)(C) (3x+5)(3x+5)(D) (x−1)(x−6)

Answer

(A)(x+5)(x−5)=x

(A)(x+5)(x−5)=x 2

(A)(x+5)(x−5)=x 2 −5

(A)(x+5)(x−5)=x 2 −5 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5)

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x)

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b)

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2 +25+30x

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2 +25+30x(D)(x−1)(x−6)=x

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2 +25+30x(D)(x−1)(x−6)=x 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2 +25+30x(D)(x−1)(x−6)=x 2 −6x−x+6

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2 +25+30x(D)(x−1)(x−6)=x 2 −6x−x+6=x

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2 +25+30x(D)(x−1)(x−6)=x 2 −6x−x+6=x 2

(A)(x+5)(x−5)=x 2 −5 2 [∵a 2 −b 2 =(a−b)(a+b)]=x 2 −25(B)(x−5)(x+5)=x 2 −5 2 =x 2 −25(C)(3x+5)(3x+5)=(3x+5) 2 (3x) 2 +5 2 +2.3x.5 [∵(a+b) 2 =a 2 +b 2 +2ab]=9x 2 +25+30x(D)(x−1)(x−6)=x 2 −6x−x+6=x 2 −7x+6