17.(6分)分解因式:
(1)$(x^2 + 2)^2 - 6(x^2 + 2) + 9$;
(2)$a^2(x - y) + 4(y - x)$.
(1)$(x^2 + 2)^2 - 6(x^2 + 2) + 9$;
(2)$a^2(x - y) + 4(y - x)$.
答案
(1) 原式$=(x^2 + 2)^2 - 6(x^2 + 2) + 9$
$=(x^2 + 2 - 3)^2$
$=(x^2 - 1)^2$
$=(x + 1)^2(x - 1)^2$
(2) 原式$=a^2(x - y) + 4(y - x)$
$=a^2(x - y) - 4(x - y)$
$=(x - y)(a^2 - 4)$
$=(x - y)(a + 2)(a - 2)$
$=(x^2 + 2 - 3)^2$
$=(x^2 - 1)^2$
$=(x + 1)^2(x - 1)^2$
(2) 原式$=a^2(x - y) + 4(y - x)$
$=a^2(x - y) - 4(x - y)$
$=(x - y)(a^2 - 4)$
$=(x - y)(a + 2)(a - 2)$
18.(6分)分解因式:
(1)$(x - 4)(x - 2) + 1$;
(2)$a^2 - 16 + 8b - b^2$.
(1)$(x - 4)(x - 2) + 1$;
(2)$a^2 - 16 + 8b - b^2$.
答案
(1)
$\begin{aligned}&(x - 4)(x - 2) + 1\\=&x^2 - 2x - 4x + 8 + 1\\=&x^2 - 6x + 9\\=&(x - 3)^2\end{aligned}$
(2)
$\begin{aligned}&a^2 - 16 + 8b - b^2\\=&a^2 - (b^2 - 8b + 16)\\=&a^2 - (b - 4)^2\\=&(a + b - 4)(a - b + 4)\end{aligned}$
$\begin{aligned}&(x - 4)(x - 2) + 1\\=&x^2 - 2x - 4x + 8 + 1\\=&x^2 - 6x + 9\\=&(x - 3)^2\end{aligned}$
(2)
$\begin{aligned}&a^2 - 16 + 8b - b^2\\=&a^2 - (b^2 - 8b + 16)\\=&a^2 - (b - 4)^2\\=&(a + b - 4)(a - b + 4)\end{aligned}$
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