MATH 150
Quiz Key #2
9/15-17/2015
(1) Use long division to calculate x5 − 7x3 + 5x + 7 ÷ x2 + 3x .
x2
+ 3x + 0
x3
x5
− x5
−3x2
+0x4
−3x4
−3x4
3x4
+2x
−7x3
+0x3
−7x3
+9x3
2x3
−2x3
−6
+0x2
+0x2
+0x2
+0x2
−6x2
−6x2
6x2
Remainder
Thus,
+5x
+7
+5x
+0x
+5x
+18x
23x
+7
+0
+7
23x + 7
x5 − 7x3 + 5x + 7 ÷ x2 + 3x = x3 − 3x2 + 2x − 6 + 2
.
x + 3x
(2) Perform the operations and simplify. List any restrictions on the values of
x.
3
3x5 + 75x3
x +8 2
÷
3x
+
11x
−
20
·
.
x2
x3 − 4x
Solution: We will first determine restrictions on the value of x. No denominator can be 0, and we can never divide by 0. Therefore,
x 6= 0 (because that
would make x2 = 0. Also, x3 − 4x = x x2 − 4 = x ( x + 2) ( x − 2) . Therefore, x 6= 2 and x 6= −2. Finally, 3x2 + 11x − 20 6= 0. But 3x2 + 11x − 20 =
(3x − 4) ( x + 5) . Therefore, x 6= 43 and x 6= −5. Therefore, the entire list of
restrictions on x is given by x 6= 0, 2, −2, 34 , −5. Now we simplify as follows:
3x3 x2 + 25
x3 + 8
1
=
·
·
x ( x + 2) ( x − 2)
(3x − 4) ( x + 5)
x2
3x2 x2 + 25
( x + 2) x2 − 2x + 4
= 2
·
( x + 2) ( x − 2)
x (3x − 4) ( x + 5)
3x2 ( x + 2) x2 + 25 x2 − 2x + 4
= 2
.
x (3x − 4) ( x + 5) ( x + 2) ( x − 2)
1
Now there is a common factor of x2 ( x + 2) in the numerator and denominator,
so we continue to simplify as:
3 x2 + 25 x2 − 2x + 4
=
(3x − 4) ( x + 5) ( x − 2)
3 x4 − 2x3 + 4x2 + 25x2 − 50x + 100
=
(3x2 + 11x − 20) ( x − 2)
=
3x4 − 6x3 + 87x2 − 150x + 200
3x3 − 6x2 + 11x2 − 22x − 20x + 40
=
3x4 − 6x3 + 87x2 − 150x + 200
.
3x3 + 5x2 − 42x + 40
(3) Fully simplify:
5 ( x + h ) −2 − 5 ( x ) −2
.
h
Solution:
=
5
( x + h )2
5
x2
h
=
=
−
5x2
−10hx − 5h2
hx2 ( x + h)
2
5x2 −5( x2 +2hx +h2 )
=
x 2 ( x + h )2
h
− 5x2
− 10hx − 5h2 1
·
h
x 2 ( x + h )2
=
−5h (2x + h)
hx2 ( x + h)
2
=
−5 (2x + h)
x 2 ( x + h )2
.
2
(4) Answer the following questions for the complex number:
√
√
√
z = −16 − 72 − 8i + 50.
(a) Write the number in standard form.
(b) Find the absolute value.
(c) Find the conjugate.
√
2+i .
(d) Calculate z ·
Solution:
(a)
z=
√
42 ·
√
√
√
√
√ √ −1 −
22 · 32 · 2 − 8i +
52 · 2
√
√
√
= 4i − 6 2 − 8i + 5 2 = − 2 − 4i.
(b)
r
|z| =
√ 2
√
√
√
√
√
− 2 + (−4)2 = 2 + 16 = 18 = 32 · 2 = 3 2.
√
(c) z = − 2 + 4i.
(d)
√
√
√
√
√
z·
2 + i = − 2 − 4i
2 + i = −2 − i 2 − 4i 2 − 4i2
√
√
= −2 − 5i 2 − 4 (−1) = 2 − 5i 2.
3