2016猿辅导初中数学竞赛训练营作业题解答-4

Posted 2020-08-15 赵胤数学课

 

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用待定系数法分解因式(1-6题)

 

1. $x^2 + xy - 2y^2 + 2x + 7y - 3$

解答: $$x^2 + xy - 2y^2 + 2x + 7y - 3 = (x + 2y + a)(x - y + b)$$ $$= x^2 + xy - 2y^2 + (a + b)x + (2b - a)y + ab$$ $$\\Rightarrow \\begin{cases}a+b = 2\\\\ 2b - a = 7\\\\ ab = -3 \\end{cases}\\Rightarrow \\begin{cases}a = -1\\\\ b =3 \\end{cases}$$ $$\\Rightarrow x^2 + xy - 2y^2 + 2x + 7y - 3 = (x - y + 3)(x + 2y - 1).$$

 

2. $a^2 - 3b^2 - 8c^2 + 2ab + 2ac + 14bc$

解答: $$a^2 - 3b^2 - 8c^2 + 2ab + 2ac + 14bc = (a + 3b + mc)(a - b + nc)$$ $$= a^2 - 3b^2 + 2ab + (m + n)ac + (3n - m)bc + mnc^2$$ $$\\Rightarrow \\begin{cases}m + n = 2\\\\ 3n - m = 14\\\\ mn = -8 \\end{cases}\\Rightarrow \\begin{cases}m = -2\\\\ n = 4\\end{cases}$$ $$\\Rightarrow a^2 - 3b^2 - 8c^2 + 2ab + 2ac + 14bc = (a + 3b - 2c)(a - b + 4c).$$

 

3. $2x^2 - 5xy - 3y^2 + 3x + 5y - 2$

解答: $$2x^2 - 5xy - 3y^2 + 3x + 5y - 2 = (2x + y + a)(x - 3y + b)$$ $$= 2x^2 - 5xy - 3y^2 + (a + 2b)x + (b - 3a)y + ab$$ $$\\Rightarrow \\begin{cases}a + 2b = 3\\\\ b - 3a = 5\\\\ ab = -2 \\end{cases} \\Rightarrow \\begin{cases}a = -1\\\\ b = 2 \\end{cases}$$ $$\\Rightarrow 2x^2 - 5xy - 3y^2 + 3x + 5y - 2 = (2x + y - 1)(x - 3y + 2).$$

 

4. $2x^2 + 3xy - 2y^2 - 5x + 5y - 3$

解答: $$2x^2 + 3xy - 2y^2 - 5x + 5y - 3 = (2x - y + a)(x + 2y + b)$$ $$= 2x^2 + 3xy - 2y^2 + (a + 2b)x + (2a - b)y + ab$$ $$\\Rightarrow \\begin{cases}a + 2b = -5\\\\ 2a - b = 5\\\\ ab = -3\\end{cases}$$ $$\\Rightarrow a = 1\\\\ b = -3$$ $$\\Rightarrow 2x^2 + 3xy - 2y^2 - 5x + 5y - 3 = (2x - y + 1)(x + 2y - 3).$$

 

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: $$x^4 + 4x^3 + 3x^2 + 4x + 2 = (x^2 + ax + b)(x^2 + cx + d)$$ $$= x^4 + (a+c)x^3 + (b + d + ac)x^2 + (ad + bc)x + bd$$ $$\\Rightarrow \\begin{cases}a + c = 4\\\\ b + d + ac = 3\\\\ ad + bc = 4\\\\ bd = 2 \\end{cases}\\Rightarrow \\begin{cases}b = \\pm1\\\\ d = \\pm2 \\end{cases}.$$ 验证 $b = 1$, $d = 2$ 可得$$\\begin{cases}a + c = 4\\\\ ac = 0\\\\ 2a + c = 4 \\end{cases}\\Rightarrow \\begin{cases}a = 0\\\\ c = 4 \\end{cases}$$ $$\\Rightarrow x^4 + 4x^3 + 3x^2 + 4x + 2 = (x^2 + 1)(x^2 + 4x + 2).$$ 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: $$a^5 + a + 1 = (a^2 +ma + n)(a^3 + pa^2 + qa + r)$$ $$= a^5 + (m + p)a^4 + (mp + q + n)a^3 + (r + mq + np)a^2 + (mr + nq)a + nr$$ $$\\Rightarrow \\begin{cases}m + p = 0\\\\ mp + q + n = 0\\\\ r + mq + np = 0\\\\ mr + nq = 1\\\\ nr = 1 \\end{cases}\\Rightarrow \\begin{cases}n = \\pm1\\\\ r = \\pm1 \\end{cases}$$ 验证 $n = 1$, $r = 1$ 可得 $$\\begin{cases}m + p = 0\\\\ mp + q + 1 = 0\\\\ 1 + mq + p = 0\\\\ m + q = 1\\end{cases} \\Rightarrow mp + q - mq - p = 0 \\Rightarrow (m - 1)(p - q) = 0 \\Rightarrow \\begin{cases}m = 1\\\\ p = -1\\\\ q = 0 \\end{cases}$$ $$\\Rightarrow a^5 + a + 1 = (a^2 + a + 1)(a^3 - a^2 + 1).$$ 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: $$x^2 + 2xy - 8y^2 + 2x + 14y + m = (x - 2y + a)(x + 4y + b)$$ $$= x^2 + 2xy - 8y^2 + (a + b)x + (4a - 2b)y + ab$$ $$\\Rightarrow \\begin{cases}a + b = 2\\\\ 4a - 2b = 14\\\\ ab = m \\end{cases}\\Rightarrow \\begin{cases}a = 3\\\\ b = -1\\\\ m = -3\\end{cases}$$ $$\\Rightarrow x^2 + 2xy - 8y^2 + 2x + 14y + m = (x - 2y + 3)(x + 4y - 1).$$

 

 

 

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