利用x^2+(p+q)x+pq型式子的因式分解下列多项式(1)x^2+7x+10; (2)X^2-2x-8;(3)y^2-7y+12; (4)x^2+7x-18.
问题描述:
利用x^2+(p+q)x+pq型式子的因式分解下列多项式
(1)x^2+7x+10; (2)X^2-2x-8;
(3)y^2-7y+12; (4)x^2+7x-18.
答
x^2+(p+q)x+pq=(x+p)(x+q)
(1)(x+2)(x+5)
(2)(x+2)(x-4)
(3)(x-3)(x-4)
(4)(x-2)(x+9)
答
(1)x²+7x+10
10=5×2
7=5+2
∴原式=(x+5)(x+2)
(2)x²-2x-8
-8=(-4)×2
-2=-4+2
∴原式=(x-4)(x+2)
(3)y²-7y+12
12=(-4)×(-3)
-7=-4-3
∴原式=(x-4)(x-3)
(4)x²+7x-18
-18=9×(-2)
7=9-2
∴原式=(x+9)(x-2)