x* - jc
3 - lOx
2 + 4x + 24.
ng the divisors of 24 yields the root 2. Division by x - 2 yields the factorization (x - 2)(*3 + x2 -I Testing the divisors of 24 yields the root 2. Division by x - 2 yields the factorization (x - 2)(*3 + x2 -
8x - 12). It turns out that-2 is another root; division of x
3 + x
2 - 8x - 12 by x + 2 gives (x-2)(x +
2)(x
2 -x-6) = (x- 2)(x + 2)(x -3)(x + 2). So, the roots are 2, -2, and 3.
x
3 - 2x
2 + x - 2.
x
3 - 2x
2 + x - 2 = x\x -2) + x-2 = (x- 2)(x
2 + 1). Thus, the only real root is 2.
x
3 + 9x
2 + 26x + 24.
Testing the divisors of 24, reveals the root -2. Dividing by x + 2 yields the factorization (x + 2)(x2 +
Ix + 12) = (x + 2)(x + 3)(;t + 4). Thus, the roots are -2, -3, and -4.
Ar
3 -5jc-2.
-2 is a root. Dividing by x + 2 yields the factorization (x + 2)(x
2 — 2x — 1). The quadratic formula
applied to x
2 - 2x — I gives the additional roots 1 ± V2.
x
3 - 4x
2 -2x + 8.
x
3 -4x
2 -2x + 8 = x
2 (x -4) - 2(x -4) = (x- 4)(x
2 - 2). Thus, the roots are 4 and ±V2.
Establish the factorization
w" - v" = (u - v)(u"~l + u"~2v + M"~ V + • • • + uv"~2 + v"~l)w" - v" = (u - v)(u"~l + u"~2v + M"~ V + • • • + uv"~2 + v"~l)
for n = 2, 3,
Simply multiply out the right-hand side. The cross-product terms will cancel in pairs (u times the kth term of
the second factor will cancel with -v times the (k - l)st term).
Prove algebraically that a real number x has a unique cube root.
Suppose there were two cube roots, u and i>, so that u
3 = i>
3 = x, or u
3 - v
3 = 0. Then, by Problem
5.83,
Unless both u and v are zero, the factor in brackets is positive (being a sum of squares); hence the other factor
must vanish, giving u = v. If both u and v are zero, then again u = v.
If f(x) — (x + 3)(x + k), and the remainder is 16 when f(x) is divided by x ~ 1, find k.
f(x) = (x~l)q(x) + \6. Hence, /(I) = 16. But, /(I) = (1 + 3)(1 + k) =4(1 + k). So, l + k = 4,
k = 3.
If f(x) = (x + 5)(x — k) and the remainder is 28 when f(x) is divided by x — 2, find k.
f(x) = (x-2)q(x) + 2&. Hence, /(2) = 28. But, f(2) = (2 + 5)(2 - k) = 7(2 - k). So, 2 - it = 4,
k= -2.
If the zeros of a function f(x) are 3 and -4, what are the zeros of the function g(x) =/(jt/3)?
/(jt/3) = 0 if and only if x/3 = 3 or */3=-4, that is, if and only if * = 9 or x=-12.
Describe the function f(x) = |jt| + j* — 1| and draw its graph.
Case 1. jcsl. Then /(*) = * + *-1 =2x - 1. Case 2. Os * < 1. Then f(x) = x - (x ~ l)= \.
Case 3. Jt<0. Then /(*)= -x - (x - 1) = -2x + 1. So, the graph (Fig. 5-25) consists of a horizontal line
segment and two half lines.
CHAPTER 5
32
5.78
5.79
5.80
5.81
5.82
5.83
5.84
5.85
5.86
5.87
5.88
3 - lOx
2 + 4x + 24.
ng the divisors of 24 yields the root 2. Division by x - 2 yields the factorization (x - 2)(*3 + x2 -I Testing the divisors of 24 yields the root 2. Division by x - 2 yields the factorization (x - 2)(*3 + x2 -
8x - 12). It turns out that-2 is another root; division of x
3 + x
2 - 8x - 12 by x + 2 gives (x-2)(x +
2)(x
2 -x-6) = (x- 2)(x + 2)(x -3)(x + 2). So, the roots are 2, -2, and 3.
x
3 - 2x
2 + x - 2.
x
3 - 2x
2 + x - 2 = x\x -2) + x-2 = (x- 2)(x
2 + 1). Thus, the only real root is 2.
x
3 + 9x
2 + 26x + 24.
Testing the divisors of 24, reveals the root -2. Dividing by x + 2 yields the factorization (x + 2)(x2 +
Ix + 12) = (x + 2)(x + 3)(;t + 4). Thus, the roots are -2, -3, and -4.
Ar
3 -5jc-2.
-2 is a root. Dividing by x + 2 yields the factorization (x + 2)(x
2 — 2x — 1). The quadratic formula
applied to x
2 - 2x — I gives the additional roots 1 ± V2.
x
3 - 4x
2 -2x + 8.
x
3 -4x
2 -2x + 8 = x
2 (x -4) - 2(x -4) = (x- 4)(x
2 - 2). Thus, the roots are 4 and ±V2.
Establish the factorization
w" - v" = (u - v)(u"~l + u"~2v + M"~ V + • • • + uv"~2 + v"~l)w" - v" = (u - v)(u"~l + u"~2v + M"~ V + • • • + uv"~2 + v"~l)
for n = 2, 3,
Simply multiply out the right-hand side. The cross-product terms will cancel in pairs (u times the kth term of
the second factor will cancel with -v times the (k - l)st term).
Prove algebraically that a real number x has a unique cube root.
Suppose there were two cube roots, u and i>, so that u
3 = i>
3 = x, or u
3 - v
3 = 0. Then, by Problem
5.83,
Unless both u and v are zero, the factor in brackets is positive (being a sum of squares); hence the other factor
must vanish, giving u = v. If both u and v are zero, then again u = v.
If f(x) — (x + 3)(x + k), and the remainder is 16 when f(x) is divided by x ~ 1, find k.
f(x) = (x~l)q(x) + \6. Hence, /(I) = 16. But, /(I) = (1 + 3)(1 + k) =4(1 + k). So, l + k = 4,
k = 3.
If f(x) = (x + 5)(x — k) and the remainder is 28 when f(x) is divided by x — 2, find k.
f(x) = (x-2)q(x) + 2&. Hence, /(2) = 28. But, f(2) = (2 + 5)(2 - k) = 7(2 - k). So, 2 - it = 4,
k= -2.
If the zeros of a function f(x) are 3 and -4, what are the zeros of the function g(x) =/(jt/3)?
/(jt/3) = 0 if and only if x/3 = 3 or */3=-4, that is, if and only if * = 9 or x=-12.
Describe the function f(x) = |jt| + j* — 1| and draw its graph.
Case 1. jcsl. Then /(*) = * + *-1 =2x - 1. Case 2. Os * < 1. Then f(x) = x - (x ~ l)= \.
Case 3. Jt<0. Then /(*)= -x - (x - 1) = -2x + 1. So, the graph (Fig. 5-25) consists of a horizontal line
segment and two half lines.
CHAPTER 5
32
5.78
5.79
5.80
5.81
5.82
5.83
5.84
5.85
5.86
5.87
5.88