Question

4a²+b²-c²-d²+4ab+2cd factorise​

Answer 1

Answer:

(2a+b+c-d)(2a+b-c+d)

Step-by-step explanation:

$sol {}^{n}$

$4 {a}^{2} + {b}^{2} - {c}^{2} - {d}^{2} + 4ab + 2cd$

$=4 {a}^{2} +4ab + {b}^{2} - {c}^{2} + 2cd - {d}^{2}$

$= (2a) {}^{2} + 2 \times 2a \times b + {b}^{2} - ( {c}^{2} - 2cd + d {}^{2} )$

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

$=(2a + b + c - d)(2a + b - c + d)$

Answer 2

Answer:

(2a+b)(2a+b) - (c-d)(c-d)

Step-by-step explanation:

4a²+b²-c²-d²+4ab+2cd

(4a²+b²+4ab) - (c²+d²-2cd)

(2a+b)²- (c-d)²

(2a+b)(2a+b) - (c-d)(c-d)