問題
次の同類項をまとめよ. また, その多項式の次数をいえ.
解説
- (1) $8x-1+5x-10x+4$
- $8x-1+5x-10x+4$
$=(8+5-10)x-1+4$
$=$$3x+3$ - 次数; 1
- $8x-1+5x-10x+4$
- (2) $4x^{3}-2x^{2}+x-1+2x^{2}-x^{3}+6$
- $4x^{3}-2x^{2}+x-1+2x^{2}-x^{3}+6$
$=(4-1)x^{3}-(2-2)x^{2}+x+(-1+6)$
$=$$3x^{3}+x+5$ - 次数; 3
- $4x^{3}-2x^{2}+x-1+2x^{2}-x^{3}+6$
- (3) $2x^{3}-x+6x^{2}-x^{3}-2x^{2}-5+3x$
- $2x^{3}-x+6x^{2}-x^{3}-2x^{2}-5+3x$
$=(2-1)x^{3}+(6-2)x^{2}+(-1+3)x-5$
$=$$x^{3}+4x^{2}+2x-5$ - 次数; 3
- $2x^{3}-x+6x^{2}-x^{3}-2x^{2}-5+3x$
- (4) $-6x^{2}-3x+5+6x^{2}-1+5x$
- $-6x^{2}-3x+5+6x^{2}-1+5x$
$=(-6+6)x^{2}+(-3+5)x+(5-1)$
$=$$2x+4$ - 次数; 1
- $-6x^{2}-3x+5+6x^{2}-1+5x$
- (5) $3-5a^{2}-a^{4}+3a^{2}-2a^{4}+2$
- $3-5a^{2}-a^{4}+3a^{2}-2a^{4}+2$
$=(-1-2)a^{4}+(-5+3)a^{2}+(3+2)$
$=$$-3a^{4}+-2a^{2}+5$ - 次数; 4
- $3-5a^{2}-a^{4}+3a^{2}-2a^{4}+2$
- (6) $2x^{2}-2xy+3y^{3}-4x^{2}+2y^{2}-3xy$
- $2x^{2}-2xy+3y^{3}-4x^{2}+2y^{2}-3xy$
$=(2-4)x^{2}+(-2-3)xy+(3+2)y^{2}$
$=$$-2x^{2}-5xy+5y^{2}$ - 次数; 2
- $2x^{2}-2xy+3y^{3}-4x^{2}+2y^{2}-3xy$