Download Exmpls 3

Math 140
Hw 3
Worked examples of selected recommended problems.
Divide to get the quotient q(x), remainder r(x). Write in
the form: p(x) = d(x)q(x) + r(x).
2
B. x −6x−2
x+5
53
x − 11 + x+5
x + 5 ) x 2 − 6x − 2
x 2 + 5x
−11x − 2
−11x − 55
53
x 2 −6x−2
53
τ x+5 = (x − 11) + x+5 Multiplying by x+5 gives
Answer: x 2 − 6x − 2 = (x + 5)(x − 11) + 53
Factor the polynomials and find all the roots (zeros).
G. x 3 + x 2 − 7x + 5 = 0 given 1 is a root.
Since 1 is a root, divide x 3 + x 2 − 7x + 5 by x − 1.
x 3 +x 2 −7x+5
The quotient is
= x 2 + 2x − 5.
x−1
Hence x 3 + x 2 − 7x + 5 = (x − 1)(x 2 + 2x − 5)
Now we must factor x 2 + 2x − 5.
The first factors to try are x ! 1 and x ! 5. Neither work,
thus we must use the quadratic formula.
x=
=
−2! 2 2 −4(1)(−5)
−b! b 2 −4ac
=
2a
2(1)
−2! 4$6
−2!2 6
−2! 24
=
= 2
2
2
=
−2! 4+20
2
= −1 ! 6
2
Hence the factorization of x + 2x − 5 is
x 2 + 2x − 5 = [x − (−1 − 6 )][x − (−1 + 6 )]
Hence x 3 + x 2 − 7x + 5 factors into
Answer: (x − 1)[x − (−1 − 6 )][x − (−1 + 6 )]
The three roots are
Answer: 1, −1 ! 6 .
H. 3x 3 − 5x 2 − 16x + 12 = 0 given -2 is a root.
Since -2 is a root, we divide by x − (−2) = x + 2.
3x 3 −5x 2 −16x+12
= 3x 2 − 11x + 6
The quotient is
x+2
Now we must factor 3x 2 − 11x + 6
3x 2 − 11x + 6 = (3x − 2)(x − 3)
Hence 3x 3 − 5x 2 − 16x + 12 factors into
Answer: (x + 2)(3x − 2)(x − 3) = (x − (−2))(3x − 2)(x − 3)
The three roots are
Answer: −2, 23 , 3.
Write domain in interval notation.
K. (a) f(t) = t 2 − 8t + 15.
There are no divisions by 0, no square roots of negative
numbers. Hence f(t) is defined everywhere.
Answer: Domain (−∞, ∞).
(b) g(t) = 1/(t 2 − 8t + 15).
Factoring gives
1
g(t) = (t−3)(t−5)
. Thus g(t) is undefined at t=3, 5.
Answer: Domain = (−∞, 3) 4 (3, 5) 4 (5, ∞).
(c) h(t) = t 2 − 8t + 15 .
This is defined provided t 2 − 8t + 15 > 0.
By part (b), the key numbers are 3, 5.
The key intervals are (-5, 3], [3, 5], [5, 5).
We use [ ] since the inequality is > rather than >.
Now evaluate t 2 − 8t + 15 at a point in each interval.
0ι(-5,3], 0 2 − 8 $ 0 + 15 = 15 = +
4ι[3,5], 4 2 − 8 $ 4 + 15 = −1 = −
10ι[5,5), 10 2 − 8 $ 10 + 15 = 35 = +
Hence h(t) is undefined between 3 and 5.
Answer: (c) Domain = (−∞, 3] 4 [5, ∞).
(d) k(t) = 3 t 2 − 8t + 15 .
You can't take the square root of a negative number but
you can take the cube root of any number. Hence k(t) is
defined everywhere.
Answer: Domain = (−∞, ∞).
L. Let f(x) = 3x 2 . Find the following.
(c) f(x 2 ).
f(x 2 ) = 3(x 2 ) 2 = 3x 4
Answer: 3x 4 .
2
(d) [f(x)] .
[f(x)] 2 = [3x 2 ] 2 = 9x 4
Answer: 9x 4 .
(e) f(x/2).
2
Answer: 34 x 2 .
f(x/2) = 3(x/2) 2 = 3 x4
M. Let H(x) = 1 − 2x 2 . Find the following.
(f) H(x + h).
H(x + h) = 1 − 2(x + h) 2 = 1 − 2(x 2 + 2xh + h 2 )
Answer: 1 − 2x 2 − 4xh − 2h 2 .
H(x+h)−H(x)
.
h
H(x+h)−H(x)
[1−2x 2 −4xh−2h 2 ]−[1−2x 2 ]
=
h
h
1−2x 2 −4xh−2h 2 −1+2x 2
=
h
2
= −4xh−2h
= −4x − 2h
h
(h)
Answer: −4x − 2h.
O. Find the quotient
g(x) = 4x 2 .
g(x)−g(a)
x−a
=
g(x)−g(a)
x−a .
[4x 2 ]−[4a 2 ]
x−a
Answer: 4x + 4a
=
4(x 2 −a 2 )
x−a
=
4(x+a)(x−a)
x−a