Mathematics Factorization by Using Identities

a)(x^2+10x+25)-(y^2+8y+16)
= (x+5)^2 - (y+4)^2
=[(x+5)+(y+4)][(x+5)-(y+4)]
=(x+y+9)(x-y+1)

b) (4x^2+4x+1)-(y^2+6y+9)
= (2x+1)^2 - (y+3)^2
=[(2x+1)+(y+3)][(2x+1)-(y+3)]
=(2x+y+4)(2x-y-2)

c) 4x^2-12xy+9y^2-16
= (2x-3y)^2 - 4^2
= [(2x-3y)+4][(2x-3y)-4]
= (2x-3y+4)(2x-3y-4)

d) 49x^2-16y^2+8y-1
= (7x)^2 - (4y-1)^2
=(7x+4y-1)(7x-4y+1)

(a) (x^2+10x+25) - (y^2+8y+16)
= (x+5)^2 - (y+4)^2
= [(x+5) - (y+4)] [(x+5) + (y+4)]
= (x - y + 1) (x + y + 9)
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(b) (4x^2+4x+1) - (y^2+6y+9)
= (2x+1)^2 - (y+3)^2
= [ (2x+1) - (y+3) ] [ (2x+1) + (y+3) ]
= (2x - y - 2) (2x + y + 4)
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(c) 4x^2-12xy+9y^2-16
= (2x-3y)^2 - 16
= (2x-3y)^2 - 4^2
= [ (2x-3y) - 4 ] [ (2x-3y) - 4 ]
= (2x - 3y - 4) (2x - 3y - 4)
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(d) 49x^2-16y^2+8y-1
同 c part 差唔多, 自己諗下
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(e) 12x^2 - 3y^2 - 8x + 2y
唔做啦, 多得滯, 一係你分開 post 嘛
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(f) y^2-14y+48

(a) (x^2+10x+25)-(y^2+8y+16)
=[x^2 + 2(5)(x) + 5^2] - [y^2 + 2(4)(y) + 4^2]
= (x+5)^2 - (y+4)^2
= (x+5+y+4)(x+5-y-4)
= (x+y+9)(x-y+1)
(b) (4x^2+4x+1)-(y^2+6y+9)
= [(2x)^2 + 2(1)(2x) + 1^2] - [y^2 + 2(3)(y) + 3^2]
= (2x+1)^2 - (y+3)^2
= (2x+1+y+3)(2x+1-y-3)
= (2x+y+4)(2x-y-2)
(c) 4x^2-12xy+9y^2-16
=(2x)^2 - 2(2x)(3y) + (3y)^2 - 4^2
= (2x-3y)^2 - 4^2
= (2x-3y+4)(2x-3y-4)
(d) 49x^2-16y^2+8y-1
= (7x)^2 - (16y^2 - 8y + 1)
= (7x)^2 - [(4y)^2 - 2(4y)(1) + 1^2]
= (7x)^2 - (4y-1)^2
= (7x+4y-1)(7x-4y+1)
(e) 12x^2-3y^2-8x+2y
=3(4x^2 - y^2) - 2(4x - y)
= 3(2x+y)(2x-y) - 2(4x-y)
It cannot be factorized
(f) y^2-14y+48
= y^2 - 14y + 49 -1
= y^2 - 2(7)(y)+7^2 - 1^2
= (y-7)^2 - 1^2
= (y-7+1)(y-7-1)
= (y-6)(y-8)