1) find the quotient and remainder ,when 4xcube -3x+9is divisible by (2x-3). 2) factorize 2xcube-11xsquare+17x-6 using synthetic division method. 3) If x cube +2xsquare +ax+b has factors (x+1) and (x-1),then find a and b.
Answers
Answer:
3.2 The Factor Theorem and The Remainder Theorem 257
3.2 The Factor Theorem and The Remainder Theorem
Suppose we wish to find the zeros of f(x) = x
3 + 4x
2 − 5x − 14. Setting f(x) = 0 results in the
polynomial equation x
3 + 4x
2 − 5x − 14 = 0. Despite all of the factoring techniques we learned1
in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing
calculator, we get
The graph suggests that the function has three zeros, one of which is x = 2. It’s easy to show
that f(2) = 0, but the other two zeros seem to be less friendly. Even though we could use the
‘Zero’ command to find decimal approximations for these, we seek a method to find the remaining
zeros exactly. Based on our experience, if x = 2 is a zero, it seems that there should be a factor
of (x − 2) lurking around in the factorization of f(x). In other words, we should expect that
x
3 + 4x
2 − 5x − 14 = (x − 2) q(x), where q(x) is some other polynomial. How could we find such
a q(x), if it even exists? The answer comes from our old friend, polynomial division. Dividing
x
3 + 4x
2 − 5x − 14 by x − 2 gives
x
2 + 6x + 7
x−2 x
3 + 4x
2 − 5x − 14
−
x
3 −2x
2
6x
2 − 5x
−
6x
2 −12x)
7x − 14
− (7x −14)
0
As you may recall, this means x
3 + 4x
2 − 5x − 14 = (x − 2)
x
2 + 6x + 7
, so to find the zeros of f,
we now solve (x − 2)
x
2 + 6x + 7
= 0. We get x − 2 = 0 (which gives us our known zero, x = 2)
as well as x
2 + 6x + 7 = 0. The latter doesn’t factor nicely, so we apply the Quadratic Formula to
get x = −3 ±
√
2. The point of this section is to generalize the technique applied here. First up is
a friendly reminder of what we can expect when we divide polynomials.
1
and probably forgot
2pun inte
258 Polynomial Functions
Theorem 3.4. Polynomial Division: Suppose d(x) and p(x) are nonzero polynomials where
the degree of p is greater than or equal to the degree of d. There exist two unique polynomials,
q(x) and r(x), such that p(x) = d(x) q(x) + r(x), where either r(x) = 0 or the degree of r is
strictly less than the degree of d.
As you may recall, all of the polynomials in Theorem 3.4 have special names. The polynomial p
is called the dividend; d is the divisor; q is the quotient; r is the remainder. If r(x) = 0 then
d is called a factor of p. The proof of Theorem 3.4 is usually relegated to a course in Abstract
Algebra,3 but we can still use the result to establish two important facts which are the basis of the
rest of the chapter.
Theorem 3.5. The Remainder Theorem: Suppose p is a polynomial of degree at least 1
and c is a real number. When p(x) is divided by x − c the remainder is p(c).
The proof of Theorem 3.5 is a direct consequence of Theorem 3.4. When a polynomial is divided
by x − c, the remainder is either 0 or has degree less than the degree of x − c. Since x − c is degree
1, the degree of the remainder must be 0, which means the remainder is a constant. Hence, in
either case, p(x) = (x − c) q(x) + r, where r, the remainder, is a real number, possibly 0. It follows
that p(c) = (c − c) q(c) + r = 0 · q(c) + r = r, so we get r = p(c) as required. There is one last ‘low
hanging fruit’4
to collect which we present below.
Theorem 3.6. The Factor Theorem: Suppose p is a nonzero polynomial. The real number
c is a zero of p if and only if (x − c) is a factor of p(x).
The proof of The Factor Theorem is a consequence of what we already know. If (x − c) is a factor
of p(x), this means p(x) = (x − c) q(x) for some polynomial q. Hence, p(c) = (c − c) q(c) = 0, so c
is a zero of p. Conversely, if c is a zero of p, then p(c) = 0. In this case, The Remainder Theorem
tells us the remainder when p(x) is divided by (x − c), namely p(c), is 0, which means (x − c) is a
factor of p. What we have established is the fundamental connection between zeros of polynomials
and factors of polynomials.
Of the things The Factor Theorem tells us, the most pragmatic is that we had better find a more
efficient way to divide polynomials by quantities of the form x − c. Fortunately, people like Ruffini
and Horner have already blazed this trail. Let’s take a closer look at the long division we performed
at the beginning of the section and try to streamline it. First off, let’s change all of the subtractions
into additions by distributing through the −1s.Step-by-step explanation: