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ÀμöÁ¤¸®
f(¥á)=0 ¢¢ f(x)=(x-¥á)Q(x) |
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f(¥á)=0
ÀÇ ¶æ
x-¥á´Â f(x)ÀÇ
ÀμöÀÌ´Ù.¡¡ |
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Ưº°ÇÑ 4Â÷½ÄÀÇ ÀμöºÐÇØ

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Problem 4-6 ¡æ
¹®Á¦¸¦ ´©¸£¸é Ç®ÀÌ¿Í ´äÀÌ ³ª¿É´Ï´Ù.
¡Ø ´ÙÀ½ ½ÄÀ» ÀμöºÐÇØ ÇϽÿÀ.
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x3+2x2-5x-6
(´ä) (x+1)(x+3)(x-2)
f(x)=x3+2x2-5x-6·Î
³õ°í »ó¼öÇ× 6ÀÇ ¾à¼ö¸¦
´ëÀÔÇØº»´Ù.
f(-1)=0À̹ǷΠx+1Àº
f(x)ÀÇ ¾à¼ö
f(x)=(x+1)(x2+x-6)=(x+1)(x+3)(x-2)
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(x2+5x+6)(x2+7x+6)-3x2
(´ä) (x2+4x+6)(x2+8x+6)
x2+6=t¶ó°í
ġȯÇϸé
ÁؽÄ=(t+5x)(t+7x)-3x2
=t2+12xt+32x2
=(t+4x)(t+8x)
=(x2+6+4x)(x2+6+8x)
=(x2+4x+6)(x2+8x+6)
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(x2+4x+3)(x2+12x+35)+15
(´ä) (x2+8x+10)(x+2)(x+6)
ÁؽÄ=(x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15 ¡ç
x2+8x¸¦ t·Î ġȯ
=(t+7)(t+15)+15
=t2+22t+120
=(t+10)(t+12)
=(x2+8x+10)(x2+8x+12)=(x2+8x+10)(x+2)(x+6)
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x4+2x3+3x2+2x+1
(´ä) (x2+x+1)2
¡¡
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