Polynomials & Algebra

(1)

please check if there are any typos.


(2)

f(x) = (x - 2)Q(x) - 2 and p(x) = (x - 2)G(x) + 2
f(x) + p(x) = (x - 2)Q(x) - 2 + (x - 2)G(x) + 2 = (x - 2)[Q(x) + G(x)]
thus, x - 2 is a factor of f(x) + p(x).

f(x) p(x) = [(x - 2)Q(x) - 2][(x - 2)G(x) + 2]
the remainder = f(2) p(2) = [(2 - 2)Q(2) - 2][(2 - 2)G(2) + 2] = -4


(3)

please check if there are any typos.


(4)(a)

2x² + x + 3 < 0
2(x² + 0.5x) + 3 < 0
2(x² + 0.5x + 0.25² - 0.25²) + 3 < 0
2(x + 0.5)² + 2.875 < 0
Since (x + 0.5)² >= 0 and so 2(x + 0.5)² + 2.875 must be greater than 0
thus, there are no real solutions.

(4)(b)

(x - 1)(x + 3) > (x - 1)(2 - x)
(x - 1)(x + 3) - (x - 1)(2 - x) > 0
(x - 1)[(x + 3) - (2 - x)] > 0
(x - 1)(2x + 1) > 0
x < -1/2 or x > 1

(4)(c)

0.9^x > 0.5
x log 0.9 > log 0.5
x < log 0.5 / log 0.9 -------------------------> (since log 0.9 is negative)
x < 6.58


(5)(a)

(x - 1)(x - 3) = (k - 1)(k - 3)
x² - 4x + 3 - k² + 4k - 3 = 0
(x + k)(x - k) - 4(x - k) = 0
(x - k)(x + k - 4) = 0
x = k or x = 4 - k

(5)(b)

(x + a)(x + b) = (c - x)(x + b)
(x + a)(x + b) - (c - x)(x + b) = 0
(x + b)[(x + a) - (c - x)] = 0
(x + b)(2x + a - c) = 0
x = -b or x = (c - a)/2

(5)(c)

(x - a)² = (2x + b)²
(x - a)² - (2x + b)² = 0
[(x - a) + (2x + b)][(x - a) - (2x + b)] = 0
(3x - a + b)(-x - a - b) = 0
x = (a - b)/3 or x = - a - b

(1)(a) remainder=2 in both cases
(b) f(y)÷(y+1) remainder=-4
Put y=x+1⇒f(x+1)÷[(x+1)+1] remainder=-4
so s(x)=x+2
(c) g(x)=Q(x)(x-1)(x+1)
g(2x+1)=Q(2x+1)(2x)(2x+2)=4Q(2x+1)x(x+1)
So common factor is x+1

2014-07-17 22:33:35 補充：
(3)(a)Let P(x)=T(x)(x+2)(x+3)-2x-7
Let Q(x)=S(x)(x+2)(x+3)
Q(x)=P(x)-(ax+b)
⇒T(x)(x+2)(x+3)-2x-7-(ax+b)=S(x)(x+2)(x+3)
So ax+b=-2x-7 hence a=-2;b=-7
(b)Q(-2x+1)=S(-2x+1)(-2x+1+2)(-2x+1+3)
=S(-2x+1)(-2x+3)(-2x+4)
So 2 factors are-2x+3 and-2x+4