扫描以下二维码下载并安装猿辅导App, 打开后请搜索教师姓名"赵胤"即可报名本课程(14次课, 99元).
用待定系数法分解因式(1-6题)
1. $x^2 + xy - 2y^2 + 2x + 7y - 3$
解答: $$x^2 + xy - 2y^2 + 2x + 7y - 3 = (x + 2y + a)(x - y + b)$$ $$= x^2 + xy - 2y^2 + (a + b)x + (2b - a)y + ab$$ $$Rightarrow egin{cases}a+b = 2\ 2b - a = 7\ ab = -3 end{cases}Rightarrow egin{cases}a = -1\ b =3 end{cases}$$ $$Rightarrow x^2 + xy - 2y^2 + 2x + 7y - 3 = (x - y + 3)(x + 2y - 1).$$
2. $a^2 - 3b^2 - 8c^2 + 2ab + 2ac + 14bc$
解答: $$a^2 - 3b^2 - 8c^2 + 2ab + 2ac + 14bc = (a + 3b + mc)(a - b + nc)$$ $$= a^2 - 3b^2 + 2ab + (m + n)ac + (3n - m)bc + mnc^2$$ $$Rightarrow egin{cases}m + n = 2\ 3n - m = 14\ mn = -8 end{cases}Rightarrow egin{cases}m = -2\ n = 4end{cases}$$ $$Rightarrow a^2 - 3b^2 - 8c^2 + 2ab + 2ac + 14bc = (a + 3b - 2c)(a - b + 4c).$$
3. $2x^2 - 5xy - 3y^2 + 3x + 5y - 2$
解答: $$2x^2 - 5xy - 3y^2 + 3x + 5y - 2 = (2x + y + a)(x - 3y + b)$$ $$= 2x^2 - 5xy - 3y^2 + (a + 2b)x + (b - 3a)y + ab$$ $$Rightarrow egin{cases}a + 2b = 3\ b - 3a = 5\ ab = -2 end{cases} Rightarrow egin{cases}a = -1\ b = 2 end{cases}$$ $$Rightarrow 2x^2 - 5xy - 3y^2 + 3x + 5y - 2 = (2x + y - 1)(x - 3y + 2).$$
4. $2x^2 + 3xy - 2y^2 - 5x + 5y - 3$
解答: $$2x^2 + 3xy - 2y^2 - 5x + 5y - 3 = (2x - y + a)(x + 2y + b)$$ $$= 2x^2 + 3xy - 2y^2 + (a + 2b)x + (2a - b)y + ab$$ $$Rightarrow egin{cases}a + 2b = -5\ 2a - b = 5\ ab = -3end{cases}$$ $$Rightarrow a = 1\ b = -3$$ $$Rightarrow 2x^2 + 3xy - 2y^2 - 5x + 5y - 3 = (2x - y + 1)(x + 2y - 3).$$
5. $x^4 + 4x^3 + 3x^2 + 4x + 2$
解答:
令原式 $=f(x)$, 验证 $f(-1)$, $f(1)$, $f(-2)$, $f(2)$, 均不等于 $0$. 因此其不含有一次因式, 即为两个二次因式乘积形式: $$x^4 + 4x^3 + 3x^2 + 4x + 2 = (x^2 + ax + b)(x^2 + cx + d)$$ $$= x^4 + (a+c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd$$ $$Rightarrow egin{cases}a + c = 4\ b + d + ac = 3\ ad + bc = 4\ bd = 2 end{cases}Rightarrow egin{cases}b = pm1\ d = pm2 end{cases}.$$ 验证 $b = 1$, $d = 2$ 可得$$egin{cases}a + c = 4\ ac = 0\ 2a + c = 4 end{cases}Rightarrow egin{cases}a = 0\ c = 4 end{cases}$$ $$Rightarrow x^4 + 4x^3 + 3x^2 + 4x + 2 = (x^2 + 1)(x^2 + 4x + 2).$$ 另解: $$x^4 + 4x^3 + 3x^2 + 4x + 2 = (x^2 + 1)^2 + 4x^3 + x^2 + 4x + 1$$ $$= (x^2 + 1)^2 + 4x(x^2 + 1) + (x^2 + 1) = (x^2 + 1)(x^2 + 4x + 2).$$
6. $a^5 + a + 1$
解答:
令原式 $= f(a)$, 验证 $f(-1)$, $f(1)$, 均不为 $0$. 因此 $f(a)$ 不含有一次因式, 即只能分解为一个二次因式与一个三次因式之乘积: $$a^5 + a + 1 = (a^2 +ma + n)(a^3 + pa^2 + qa + r)$$ $$= a^5 + (m + p)a^4 + (mp + q + n)a^3 + (r + mq + np)a^2 + (mr + nq)a + nr$$ $$Rightarrow egin{cases}m + p = 0\ mp + q + n = 0\ r + mq + np = 0\ mr + nq = 1\ nr = 1 end{cases}Rightarrow egin{cases}n = pm1\ r = pm1 end{cases}$$ 验证 $n = 1$, $r = 1$ 可得 $$egin{cases}m + p = 0\ mp + q + 1 = 0\ 1 + mq + p = 0\ m + q = 1end{cases} Rightarrow mp + q - mq - p = 0 Rightarrow (m - 1)(p - q) = 0 Rightarrow egin{cases}m = 1\ p = -1\ q = 0 end{cases}$$ $$Rightarrow a^5 + a + 1 = (a^2 + a + 1)(a^3 - a^2 + 1).$$ 另解: $$a^5 + a + 1 = a^5 - a^2 + a^2 + a + 1$$ $$= a^2(a - 1)(a^2 + a + 1)+ (a^2 + a + 1) = (a^2 + a + 1)(a^3 - a^2 + 1).$$
7. 确定 $m$ 的值, 使 $x^2 + 2xy - 8y^2 + 2x + 14y +m$ 能分解为两个一次式的积.
解答: $$x^2 + 2xy - 8y^2 + 2x + 14y + m = (x - 2y + a)(x + 4y + b)$$ $$= x^2 + 2xy - 8y^2 + (a + b)x + (4a - 2b)y + ab$$ $$Rightarrow egin{cases}a + b = 2\ 4a - 2b = 14\ ab = m end{cases}Rightarrow egin{cases}a = 3\ b = -1\ m = -3end{cases}$$ $$Rightarrow x^2 + 2xy - 8y^2 + 2x + 14y + m = (x - 2y + 3)(x + 4y - 1).$$
赵胤老师微信二维码:
