How To Do Synthetic Division To Find Zeros

Roots of a Polynomial

A root or nix of a role is a number that, when plugged in for the variable, makes the role equal to nothing. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0.

The Rational Zeros Theorem

The Rational Zeros Theorem states:

If P(x) is a polynomial with integer coefficients and if is a nil of P(x) ( P() = 0), and so p is a cistron of the constant term of P(x) and q is a factor of the leading coefficient of P(x).

Nosotros can apply the Rational Zeros Theorem to find all the rational zeros of a polynomial. Here are the steps:

Adapt the polynomial in descending order
Write downward all the factors of the constant term. These are all the possible values of p .
Write down all the factors of the leading coefficient. These are all the possible values of q .
Write downward all the possible values of . Remember that since factors can exist negative, and - must both be included. Simplify each value and cantankerous out any duplicates.
Use synthetic sectionalization to determine the values of for which P() = 0. These are all the rational roots of P(x).

Example: Find all the rational zeros of P(x) = x ³ -ninex + 9 + iix ^iv -19ten ² .

P(ten) = 2x ⁴ + ten ³ -xixx ⁱⁱ - 910 + 9
Factors of constant term: ±ane, ±3, ±9.
Factors of leading coefficient: ±i, ±2.
Possible values of : ± , ± , ± , ± , ± , ± . These can be simplified to: ±ane, ± , ±3, ± , ±9, ± .
Use synthetic division:

Thus, the rational roots of P(x) are x = - 3, -1, , and three.

We can ofttimes use the rational zeros theorem to factor a polynomial. Using synthetic division, we can find one real root a and we can notice the quotient when P(x) is divided past x - a . Next, nosotros can use synthetic division to find one factor of the quotient. We can go on this procedure until the polynomial has been completely factored.

Example (as above): Factor P(x) = 2x ⁴ + x ³ -19x ² - ninex + 9.

Every bit seen from the second constructed division above, twoten ⁴ + ten ^three -19x ² -ixx + ix÷x + one = 2x ³ - x ⁱⁱ - 18ten + 9. Thus, P(10) = (ten + 1)(2ten ³ - ten ^two - eighteenx + 9). The second term can be divided synthetically by 10 + 3 to yield 2x ² - vii10 + three. Thus, P(x) = (ten + 1)(x + 3)(2x ² - 7x + iii). The trinomial can and then be factored into (x - 3)(2ten - ane). Thus, P(10) = (x + 1)(10 + 3)(x - iii)(ii10 - 1). We can see that this solution is correct because the 4 rational roots constitute in a higher place are zeros of our result.