Answers to the Above Questions

.
P(x)Q(x) = (2x^{ 2}  3x)(3x^{ 2} + x  5)
= 6x^{ 4}  7x^{ 3} 13x^{ 2} + 15x

.
P(x)/ Q(x) = (2x^{ 4}  x^{ 3} 3x^{ 2} + 7x  13) / (x^{ 2}  4)
= 2x^{ 2}  x + 5 + (3x + 7) / (x^{ 2}  4)
quotient = 2x^{ 2}  x + 5 , remainder = 3x + 7

Zeros at 0, 1, 3 and 4, hence P(x) can be written as
P(x) = A x (x  1)(x  3)(x  4)
We now use the point (2 , 2) to write P(2) = 2 and use it to find the constant A = 1/2.
P(x) = (1/2) x (x  1)(x  3)(x  4)
Expand to find
P(x) = (1/2)x^{ 4}  4x^{ 3} + (19/2)x^{ 2}  6x

Using remainder theorem,
remainder = P(1) = 4*1^{ 200} + 5*1^{ 95}  4*1^{ 21} + 2*1  6 = 1

P(x) = 2(x  1)(x + 2)(x  3) = 2 x^{ 3} + 4 x^{ 2} + 10 x  12

According to the rational zeros theorem, the possible rational zeros are as follows: ~+mn~1 , ~+mn~1/3 , ~+mn~5/3 , ~+mn~3 , ~+mn~5
Substituting or using synthetic division, it can easily be shown that 3 and 5/3 are rational zeros of P(x). hence P(x) may be written as
P(x) = (x + 3)(x + 5/3) Q(x)
Q(x) = P(x) / (x + 3)(x + 5/3) = 3x^{ 2} + 12 x + 3 (using division)
The remaining zeros of P(x) are the zeros of Q(x) which are 2 + √3 and 2  √3.

If (x  2) is a factor, then P(2) = 0. The remainder theorem also states that P(1) = 3. After substitutions P(2) and P(1), we obtain a system of 2 equations in c and d.
2c + d = 12 and c + d = 1
solve to find: c = 13 and d = 14

If P(2) = 0, then 2 is a zero of P. Since P is even 2 is also a zero of P because of the symmetry of the graph of P with respect to y axis.
If P(x + 3) has a zero at x = 4, then P(x) has a zero at x = 1 since the graph of P(x + 3) is the graph of P(x) shifted 3 units left. Also and because of symmetry x = 1 is a zero of P. Hence P has 4 zeros: 2, 1, 1 and 2 and P(0) = 9.
P(x) = A (x + 1)(x + 2)(x  1)(x  2)
Use P(0) = 9 to find coefficient A = 9/4
P(x) = (9/4) (x + 1)(x + 2)(x  1)(x  2) = (9/4) x^{ 4} + (45/4)x^{ 2}  9

Since x = 1/2 and x = 3 are zeros of P, we can write
P(x) = (x  1/2)(x + 3) Q(x)
Q(x) = P(x) / (x  1/2)(x + 3) = 2(x^{ 2}  x  4) (using division)
The remaining zeros of P(x) are the zeros of Q(x): 1/2 + (1/2)√17 and 1/2  (1/2)√17.

a) f(x) = x^{ 3}  x^{ 2}  4 x + 4
= x^{ 2}(x  1)  4(x  1)
= (x  1)(x^{ 2}  4)
= (x  1)(x  2)(x + 2)
b) f(x) = 2 x^{ 2}  x^{ 3}
= x^{ 2}(2  x)
c) f(x) = x^{ 2}  2 x^{ 4}
= x^{ 2}(1  2x^{ 2})
= x^{ 2}(1  x&radic2)(1 + x&radic2)
d) f(x) = (x^{ 2}  2 x  3)^{ 2}
= ((x + 1)(x  3))^{ 2}
= (x + 1)^{ 2}(x  3)^{ 2}
e) f(x) = x^{ 4} + 3 x^{ 3} + 3 x^{ 2} + x
= x (x^{ 3} + 3 x^{ 2} + 3 x + 1)
= x(x + 1)^{ 3}

Leading coefficient negative therefore between x = 2 and x = 1 graph is above x axis and below x axis between x = 1 and x = 1 and hence the y intercept is below the x axis.

1  C
2  D
3  B
4  A

1) P is even , the graph is not symmetric with respect to y axis.
2) P has no zeros, graph has x intercepts.
3) P(0) = 1 , y intercept in graph below x axis.
4) Leading coefficient in P is negative, graph going down on the left and on the right.

a) The degree of f odd since its graph has 5 xintercepts and the complex zeros come in pairs.
b) The leading coefficient is negative because the graph is going down on the right and up on the left.
c) No, the degree of a polynomial is determined by real and complex zeros. Only the real zeros in the form of xintercepts are shown on the graph.

The given polynomial, the divisor (x^{2}  1), the remainder 5x + 4 and the quotient Q(x) are related as follows:
3x^{3} + qx^{2} + px + 7 = Q(x)(x^{2}  1) + 5x + 4.
Substitute x by 1 and 1 on both sides of the equation to obtain 2 equations in q and p.
x = 1: 3 + q + p + 7 = 0 + 9 (1)
x = 1: 3 + q  p + 7 = 1
Solve the above system of equations to find: q = 3 and p = 2.
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