問題
次の式を展開せよ.
解説
- (1) $2x(x^{2}+x+5)+4(1-4x-x^{2})-x(5x-4+3x^{2})$
- $2x(x^{2}+x+5)+4(1-4x-x^{2})-x(5x-4+3x^{2})$
$=2x^{3}+2x^{2}+10x+4-16x-4x^{2}-5x^{2}+4x-3x^{3}$
$=$$-x^{3}-7x^{2}-2x+4$
- $2x(x^{2}+x+5)+4(1-4x-x^{2})-x(5x-4+3x^{2})$
- (2) $(a+b)(a^{3}-a^{2}b+ab^{2}-b^{3})$
- 既に因数分解を習っていれば,
$(a+b)(a^{3}-a^{2}b+ab^{2}-b^{3})$
$=(a+b)\{a^{2}(a-b)+b^{2}(a-b)\}$
$=(a+b)(a-b)(a^{2}+b^{2})$
$=(a^{2}-b^{2})(a^{2}+b^{2})$
$=$$a^{4}-b^{4}$ - 一般に,
$a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\cdots+a^{2}b^{n-2}+b^{n-1})$
$=$$(a-b)\cdot\displaystyle\sum_{k=1}^{n-1}a^{n-k}b^{k}$
と因数分解ができます.
- 既に因数分解を習っていれば,
- (3) $(3x-2x^{2}-4)(x^{2}+5-3x)$
- $(3x-2x^{2}-4)(x^{2}+5-3x)=3x^{3}+15x-9x^{2}-2x^{4}-10x^{2}+6x^{3}-4x^{2}-20+12x$
$=$$-2x^{4}+9x^{3}-23x^{2}+27x-20$
- $(3x-2x^{2}-4)(x^{2}+5-3x)=3x^{3}+15x-9x^{2}-2x^{4}-10x^{2}+6x^{3}-4x^{2}-20+12x$
