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標題:

HKDSE MATHEMATICS

• 數學知識交流---考据的問題之其一_0@1@
• Zonda R@1@
• 解下列方程組

 

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發問:

f(x) = x3+ 2x2 + px + q is divisible by x2 + x -6. (a) Find the values of the constants p and q(b) Hence factorize f(x)

最佳解答:

(a) f(x)= x3+ 2x2 + px + q is divisible by x2 + x - 6. x2+ x - 6 =(x + 3)(x - 2) Hence, f(x) is divisible by x + 3. f(-3) = 0 (-3)3+ 2(-3)2 +p(-3) +q =0 -27 + 18 - 3p + q = 0 -3p + q = 9 ...... (1) f(x) is also divisible by x - 2. f(2) = 0 (2)3+ 2(2)2 +p(2) +q =0 8 + 8 + 2p + q = 0 2p + q = -16 ...... (2) (2) - (1) : 5p = -25 p = -5 Put p = -5 into (2) : 2(-5) + q = -16 q = -6 Hence, p = -5 and q = -6 ========== (b) f(x) = x3+2x2 - 5x - 6 By long division, f(x) ÷ (x2 + x - 6) =(x3+2x2 - 5x - 6) ÷ (x2 + x - 6) = (x + 1) Hence, f(x) = (x2 + x - 6)(x + 1) = (x + 3)(x - 2)(x + 1)

其他解答:

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