因式分解體f.3)數學

1.(a)因式分解體x+1)(x-6)-8
(x + 1) (x - 6) - 8
= x^2 - 6x + x - 6 - 8
= x^2 - 5x -14
= (x + 2) (x - 7) (ans.)……..(1)

(b)由此,因戌分解
(2y^2+5y+1)(2y^2+5y-6)-8……..(2)
Subsitute x = (2y^2 + 5) into (2),
Thus, (2)
= (2y^2 + 5 + 2) (2y^2 + 5 - 7)
= (2y^2 + 7) (2y^2 - 2)
= 2(2y^2 + 7) (y^2 - 1)
= 2( y + 2) (y + 5) (y + 1) (y - 1) (ans.)

2.(a)因式分解
16y^2-16y+4
16y^2 - 16y + 4
= (4y)^2 - (2)(4y)(2) + (2)^2
= (4y - 2)^2……..(1)
= 4(2y - 1)^2 (ans.)



(b)由此,或用其他方法因式分解
(16x^4-16x^2+4)-49x^2 ……..(2)

Substitute x = (16y^2 - 16y + 4) into (2)
Thus, (2)
= (4y^2 - 2)^2 - 49 (4y^2 - 2)^2
= (4y^2 - 2)^2 - [7(4y^2 - 2)]^2
= [(4y^2 - 2 ) + 7(4y^2 - 2)] [(4y^2 - 2 ) - 7(4y^2 - 2)]
= [32y^2 - 16] [-24y^2 + 12]
= (16)(y^2 - 1) (12) (1 - 2y^2)
= 192(y + 1)(y - 1)(1 - 2y^2) (ans.)

希望幫到你.

1.(a)
(x + 1) (x - 6) - 8
= x^2 - 6x + x - 6 - 8
= x^2 - 5x -14
= (x + 2) (x - 7) ……..(1)
(b)
(2y^2+5y+1)(2y^2+5y-6)-8……..(2)
Subsitute x = (2y^2 + 5) into (2)
(2)
= (2y^2 + 5 + 2) (2y^2 + 5 - 7)
= (2y^2 + 7) (2y^2 - 2)
= 2(2y^2 + 7) (y^2 - 1)
= 2( 2y^2 + 7) (y + 5) (y + 1) (y - 1) 2.(a)
16y^2 - 16y + 4
= (4y)^2 - (2)(4y)(2) + (2)^2
= (4y - 2)^2……..(1)
= 4(2y - 1)^2
(b)
(16x^4-16x^2+4)-49x^2 ……..(2)
Substitute x = (16y^2 - 16y + 4) into (2)
(2)
= (4y^2 - 2)^2 - 49 (4y^2 - 2)^2
= (4y^2 - 2)^2 - [7(4y^2 - 2)]^2
= [(4y^2 - 2 ) + 7(4y^2 - 2)] [(4y^2 - 2 ) - 7(4y^2 - 2)]
= [32y^2 - 16] [-24y^2 + 12]
= (16)(y^2 - 1) (12) (1 - 2y^2)
= 192(y + 1)(y - 1)(1 - 2y^2)