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1. Polynomials
1.1. Definitions
A polynomial in x is an expression obtained by taking powers of x, multiplying them by
constants, and adding them. It can be written in the form
c0 xn + c1 xn−1 + c2 xn−2 + · · · + cn−1 x + cn
where n is an integer ≥ 0, and c0 , c1 , . . . , cn are constants. For example
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2x6 + 5x5 + x3 + x2 − 6x − 10
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Note that x 2 + 3 and 4x3 + x32 are not polynomials.
The constants c0 , . . . , cn are called the coefficients.
The constant term is cn .
The leading term is the term involving the highest power of x, here c0 xn .
The degree is the power of x in the leading term.
A degree 0 polynomial is just a constant, e.g., 2 is (a) constant
A degree 1 polynomial is called linear, e.g., 3x + 2 is linear
A degree 2 polynomial is called quadratic, e.g., x2 + 2x + 1 is quadratic
A degree 3 polynomial is called cubic, e.g., y 3 + 7y − 2 is a cubic in y.
1.2. Operations
Polynomials can be added or subtracted simply by adding or subtracting the corresponding
terms, e.g., if
f (x) = x2 − 3,
g(x) = 4x3 + 3x2 − 2x + 6
then
f (x) + g(x) = 4x3 + 4x2 − 2x + 3,
f (x) − g(x) = −4x3 − 2x2 + 2x − 9
To multiply two polynomials, every term of the first polynomial must be multiplied by
every term of the second.
Example 1.1.
f (x)g(x) =(x2 − 3)(4x3 + 3x2 − 2x + 6)
=x2 (4x3 + 3x2 − 2x + 6) − 3(4x3 + 3x2 − 2x + 6)
=4x5 + 3x4 − 2x3 + 6x2 − 12x3 − 9x2 + 6x − 18
=4x5 + 3x4 − 14x3 − 3x2 + 6x − 18
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1.3. Polynomial division
Similar to long division for ordinary numbers.
To divide 2x3 + 10x2 − 3x + 1 by x + 3, we compute
x+3|
2x2 +4x −15
2x3 + 10x2 −3x +1
2x3 + 6x2
4x2 −3x
4x2 +12x
−15x +1
−15x −45
46
The term 2x3 must come from multiplying x+3 by 2x2 , so put 2x2 on the top line, multiply
x + 3 by 2x2 and subtract from the line above.
Now bring down the −3x. Multiply x + 3 by 4x and subtract.
Bring down the −1. Multiply x + 3 by −15 and subtract.
This completes the process.
The quotient is 2x2 + 4x − 15 and the remainder is 46.
polynomial = divisor · quotient + remainder
2x3 + 10x2 − 3x + 1 = (x + 3)(2x2 + 4x − 15) + 46
Note. If the remainder is zero, we say that the polynomial is exactly divisible by the
divisor, or that the divisor is a factor of the polynomial.
Example 1.2. Divide x4 − 2x2 + 3x − 6 by x2 − 4x + 3.
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x − 4x + 3 |
4
x
x4 −4x3
4x3
4x3
x2
+4x +11
2
−2x
+3x −6
+3x2
−5x2 +3x
−16x2 +12x
11x2
−9x −6
11x2 −44x +33
35x −39
The quotient is x2 + 4x + 11. The remainder is 35x − 39.
polynomial = divisor · quotient + remainder
x4 − 2x2 + 3x − 6 = (x2 − 4x + 3)(x2 + 4x + 11) + 35x − 39
Note. It is always true that degree of remainder < degree of divisor.
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1.4. Remainder Theorem
Theorem 1.3. When a polynomial f (x) is divided by x − a, then the remainder is f (a).
Example 1.4. The remainder when f (x) = 2x5 − 3x2 + 1 is divided by x + 2 is f (−2) =
(−2)5 − 3(−2)2 + 1 = −32 − 12 + 1 = −43. Check this yourself by doing the long division.
Proof. If the quotient is q(x) and the remainder is r, then
f (x) = (x − a).q(x) + r
Substituting x = a, gives f (a) = (a − a) · q(a) + r = r.
1.5. Factor Theorem
Theorem 1.5. If x − a is a factor of a polynomial f (x), then f (a) = 0. Conversely, if
f (a) = 0, then x − a is a factor of f (x).
Example 1.6. Is x − 1 is a factor of f (x) = x5 − 1? Since f (1) = 0, the Factor Theorem
tells us it must be. In fact x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1).
Proof. If x − a is a factor of f (x) then f (x) = (x − a) · g(x) for some polynomial g(x).
Therefore f (a) = (a − a) · g(a) = 0.
If f (a) = 0 then by the Remainder Theorem, the remainder on dividing f (x) by x − a is
0. Therefore f (x) = (x − a) · q(x) where q(x) is the quotient.
1.6. Factorizing polynomials
It is often useful to write a polynomial as a product of polynomials of lower degree. For
example
f (x) = x3 − 2x2 − x + 2
can be factorized as (x2 − 1)(x − 2) or as (x − 1)(x + 1)(x − 2).
Factorizing helps find the roots since, if a product of numbers is zero, then one of
them must be zero.
Therefore if f (x) = 0 then x − 1 = 0, or x + 1 = 0 or x − 2 = 0. Therefore x = 1, −1 or 2.
Some well-known factorizations:
(i) Difference of squares:
x2 − a2 = (x − a)(x + a)
for example, x2 − 9 = (x + 3)(x − 3).
(ii) Perfect square:
x2 ± 2ax + a2 = (x ± a)2
for example x2 − 10x + 25 = (x − 5)2 .
(iii) If there is no constant term, then x is a common factor. For example,
x2 + 4x = x(x + 4).
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1.7. Factorizing by trial and error
Given a quadratic x2 + bx + c, we try to write it as
x2 + bx + c = (x + r)(x + s)
for some r, s. Comparing coefficients, we need to find two numbers r and s with r + s = b
and rs = c. If r and s are integers, they must be factors of c. Sometimes they can be
found by trial and error.
Example 1.7. x2 + 7x + 10. Need rs = 10. You can write 10 = 1 × 10 = 2 × 5. The
second one works since 2 + 5 = 7. Therefore x2 + 7x + 10 = (x + 2)(x + 5).
Example 1.8. x2 + 3x − 18. Need rs = −18, so one of the numbers r and s must be
positive, the other negative. Now 18 = 1 × 18 = 2 × 9 = 3 × 6. In fact r = 6, s = −3
works, since r + s = 3. Therefore x2 + 3x − 18 = (x + 6)(x − 3).
To factorize ax2 + bx + c with a 6= 1 by inspection is harder, but it can sometimes be done.
Example 1.9. 2x2 + 7x + 6. Try to write it as (2x + r)(x + s). Need rs = 6 and r + 2s = 7.
Take r = 3 and s = 2, so 2x2 + 7x + 6 = (2x + 3)(x + 2).
1.8. Using the Factor Theorem
If f (a) = 0 then x − a is a factor of f (x).
If f (x) = ax2 + bx + c has roots x = x1 , x2 , then f (x) = a(x − x1 )(x − x2 ).
√
−6 ± 324
Example 1.10. From the standard formula, x + 6x − 72 has roots x =
giving
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x = 6, −12. Therefore x2 + 6x − 72 = (x − 6)(x + 12).
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It can be difficult to factorize polynomials of degree ≥ 3. If you happen to know a root,
say f (a) = 0, then x − a is a factor, so f (x) = (x − a) · q(x). Use polynomial division to
find q(x), and try to factorize it.
1.9. Worked examples
Example 1.11. Find the quotient and remainder for (x3 + 5x2 + 4x − 17) ÷ (x − 4).
Example 1.12. Find the quotient and remainder for (x3 − 4x2 + x + 2) ÷ (x2 − 3).
Example 1.13. When x3 − 4x2 + 5x + a is divided by x − 3, the remainder is 7. Find a.
Example 1.14. Factorize the following polynomials:
x2 + 7x + 12,
x2 + 8x,
x3 − 5x2 − 50x,
2x2 − 5x + 3,
x3 + 3x2 − 4x − 12.
Example 1.15. The polynomial x3 + 8x2 + kx + 10 has a factor x + 2. (a) Find the value
of k. (b) Factorize the polynomial.
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Example 1.16. Given that f (x) = x3 −4x2 −x+4, find f (1) and f (2), and hence factorize
f (x) into a product of three linear factors.
Example 1.17. Express 6x3 + 17x2 + 14x + 3 in the form (x + 1)(ax2 + bx + c), stating
the values of a, b and c, and hence solve the equation 6x3 + 17x2 + 14x + 3 = 0.
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