06/14 6. Polynomial Expressions and Functions

A structured guide to operating on polynomials, dividing them, applying the Remainder and Factor Theorems, and finding polynomial zeros through factoring and rational-root testing.

Polynomial Structure and Operations

A polynomial expression is a sum of terms of the form axnax^n, where aa is a constant and nn is a nonnegative integer. A polynomial function can be written as

f(x)=anxn+an−1xn−1+⋯+a1x+a0,f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,

where an≠0a_n\neq 0. The is the greatest exponent with a nonzero coefficient. The coefficient of that highest-degree term is the , and a0a_0 is the constant term.

For example, in

f(x)=4x5−3x2+7,f(x)=4x^5-3x^2+7,

the degree is 55, the is 44, and the constant term is 77.

Combining and multiplying expressions

To add or subtract polynomials, combine like terms: terms with the same variable raised to the same power. When subtracting, distribute the negative sign first.

(3x3−2x+5)+(x3+7x−1)=4x3+5x+4(3x^3-2x+5)+(x^3+7x-1)=4x^3+5x+4
(5x2−3x+4)−(2x2+x−6)=5x2−3x+4−2x2−x+6=3x2−4x+10.\begin{aligned} (5x^2-3x+4)-(2x^2+x-6) &=5x^2-3x+4-2x^2-x+6\\ &=3x^2-4x+10. \end{aligned}

To multiply, use the distributive property so that every term in one factor is multiplied by every term in the other factor:

(x+3)(2x2−x+4)=x(2x2−x+4)+3(2x2−x+4)=2x3+5x2+x+12.\begin{aligned} (x+3)(2x^2-x+4) &=x(2x^2-x+4)+3(2x^2-x+4)\\ &=2x^3+5x^2+x+12. \end{aligned}

The degree of a product is the sum of the degrees of its nonzero polynomial factors.

Takeaway: Identify the highest power to describe a polynomial, combine only like terms, and distribute across every term when multiplying.

Polynomial Identities and Factoring

Polynomial identities are equations that hold for every value of the variables. Recognizing them makes factoring faster and reduces the amount of expansion required.

Frequently used identities

(a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2
(a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2
(a+b)(a−b)=a2−b2(a+b)(a-b)=a^2-b^2
(a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3
(a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^3
a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2)
a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2)

For example, 9x2−259x^2-25 is a difference of squares:

9x2−25=(3x)2−52=(3x−5)(3x+5).9x^2-25=(3x)^2-5^2=(3x-5)(3x+5).

When factoring, first check for a greatest common factor. Then look for a familiar identity, such as a difference of squares or a sum or difference of cubes.

Takeaway: Factoring is often the reverse of multiplication, so matching an expression to a standard identity can reveal its factors immediately.

Polynomial Division

Polynomial division follows the same structure as integer division. If a polynomial f(x)f(x) is divided by a nonzero polynomial d(x)d(x), then

f(x)=d(x)q(x)+r(x),f(x)=d(x)q(x)+r(x),

where q(x)q(x) is the quotient and the degree of r(x)r(x) is less than the degree of d(x)d(x).

Long division

To divide 2x3+3x2−5x+62x^3+3x^2-5x+6 by x+2x+2:

  1. Divide the leading terms: 2x3÷x=2x22x^3\div x=2x^2.

  2. Multiply: 2x2(x+2)=2x3+4x22x^2(x+2)=2x^3+4x^2.

  3. Subtract to obtain −x2-x^2, bring down the next term, and repeat.

  4. Continue until the remainder has lower degree than the divisor.

The result is

2x3+3x2−5x+6=(x+2)(2x2−x−3)+12.2x^3+3x^2-5x+6=(x+2)(2x^2-x-3)+12.

Thus, the quotient is 2x2−x−32x^2-x-3, and the remainder is 1212.

is a shorter method when the divisor is x−cx-c. Since x+2=x−(−2)x+2=x-(-2), use −2-2 with the coefficients 2,3,−5,62,3,-5,6:

−223−56−4262−1−312\begin{array}{r|rrrr} -2&2&3&-5&6\\ &&-4&2&6\\ \hline &2&-1&-3&12 \end{array}

The last entry is the remainder, and the preceding entries are the coefficients of the quotient. Therefore, the quotient is 2x2−x−32x^2-x-3, with remainder 1212.

Takeaway: Long division works for general polynomial divisors, while provides a compact method for linear divisors.

Remainders and Factors

The connects division with evaluation. When f(x)f(x) is divided by x−cx-c, the remainder is f(c)f(c):

f(x)=(x−c)q(x)+f(c).f(x)=(x-c)q(x)+f(c).

For

f(x)=x3−4x2+2x+7,f(x)=x^3-4x^2+2x+7,

the remainder after division by x−3x-3 is found without performing long division:

f(3)=33−4(32)+2(3)+7=27−36+6+7=4.\begin{aligned} f(3)&=3^3-4(3^2)+2(3)+7\\ &=27-36+6+7\\ &=4. \end{aligned}

Therefore,

f(x)=(x−3)q(x)+4f(x)=(x-3)q(x)+4

for some quadratic polynomial q(x)q(x).

The is a direct consequence:

x−c is a factor of f(x)  ⟺  f(c)=0.x-c\text{ is a factor of }f(x)\iff f(c)=0.

For example, let

f(x)=x3−5x2+2x+8.f(x)=x^3-5x^2+2x+8.

Evaluating at 22 gives

f(2)=8−20+4+8=0.f(2)=8-20+4+8=0.

Thus, x−2x-2 is a factor. produces the quotient x2−3x−4x^2-3x-4, so

f(x)=(x−2)(x2−3x−4)=(x−2)(x−4)(x+1).f(x)=(x-2)(x^2-3x-4)=(x-2)(x-4)(x+1).

Takeaway: Evaluate f(c)f(c) to find the remainder for division by x−cx-c; a remainder proves that x−cx-c is a factor.

Zeros and

A or root of ff is a number cc such that f(c)=0f(c)=0. The zeros are the xx-coordinates of the graph's xx-intercepts. By the , each cc corresponds to a factor x−cx-c.

Factoring to find zeros

Suppose

f(x)=x3−6x2+11x−6.f(x)=x^3-6x^2+11x-6.

Factoring gives

f(x)=(x−1)(x−2)(x−3).f(x)=(x-1)(x-2)(x-3).

Set each factor equal to :

x−1=0,x−2=0,x−3=0.x-1=0,\qquad x-2=0,\qquad x-3=0.

The zeros are x=1x=1, x=2x=2, and x=3x=3.

and graph behavior

A repeated factor gives a with greater than one. For

f(x)=(x−2)3(x+1)2,f(x)=(x-2)^3(x+1)^2,

x=2x=2 has 33, and x=−1x=-1 has 22. A with odd generally causes the graph to cross the xx-axis, while a with even generally causes the graph to touch the axis and turn around.

A polynomial of degree nn has exactly nn complex zeros when multiplicities are counted. Some may be nonreal. For example,

f(x)=x2+4f(x)=x^2+4

has no real zeros but has the two complex zeros x=2ix=2i and x=−2ix=-2i.

Takeaway: Factor the polynomial, set each factor equal to , and count repeated factors when checking the total number of complex zeros.

A Strategy for Finding Polynomial Zeros

The narrows the search for rational zeros of a polynomial with integer coefficients. If

f(x)=anxn+⋯+a1x+a0,f(x)=a_nx^n+\cdots+a_1x+a_0,

then every rational pq\frac{p}{q}, written in lowest terms, must have pp as a factor of the constant term a0a_0 and qq as a factor of the ana_n.

For

f(x)=2x3−3x2−8x+12,f(x)=2x^3-3x^2-8x+12,

the possible rational zeros are

±1, ±2, ±3, ±4, ±6, ±12, ±12, ±32.\pm1, \ \pm2, \ \pm3, \ \pm4, \ \pm6, \ \pm12, \ \pm\frac{1}{2}, \ \pm\frac{3}{2}.

A systematic strategy is:

  1. Write the polynomial in standard form.

  2. Factor out any greatest common factor.

  3. Look for identities such as a difference of squares or a sum or difference of cubes.

  4. List possible rational zeros.

  5. Test candidates by substitution or the .

  6. Use when a is found.

  7. Factor the reduced polynomial.

  8. Set every factor equal to and solve.

  9. Check the number of zeros against the degree, counting multiplicities.

For the example above, test x=2x=2:

f(2)=2(8)−3(4)−8(2)+12=16−12−16+12=0.f(2)=2(8)-3(4)-8(2)+12=16-12-16+12=0.

Therefore, x−2x-2 is a factor. gives

22−3−81242−1221−60\begin{array}{r|rrrr} 2&2&-3&-8&12\\ &&4&2&-12\\ \hline &2&1&-6&0 \end{array}

Thus,

f(x)=(x−2)(2x2+x−6)=(x−2)(2x−3)(x+2).f(x)=(x-2)(2x^2+x-6)=(x-2)(2x-3)(x+2).

The zeros are

x=2,x=32,x=−2.x=2,\qquad x=\frac{3}{2},\qquad x=-2.

Takeaway: The supplies candidates, the tests them, and reduces the problem after a candidate succeeds.