Finding Roots With the Remainder Theorem. Dividing Polynomials Remember, when we divide a polynomial by another polynomial, we get a quotient and a remainder.

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Presentation transcript:

Finding Roots With the Remainder Theorem

Dividing Polynomials Remember, when we divide a polynomial by another polynomial, we get a quotient and a remainder. In addition, when you divide a polynomial by (x – a), if your remainder is 0, a is a zero of the polynomial. This is, in fact, just one application of the remainder theorem.

The Remainder Theorem The remainder theorem states that the remainder that you get when you divide a polynomial P(x) by (x – a) is equal to P(a). We can use this to find the value of a function at a point or to find its zeroes – if the remainder of P(x) / (x – a) is zero, P(x) has a zero at a. This is often the easiest way to test if P(x) has a root at a.

Why is this? Let’s call the remainder that we get from dividing P(x) by (x – a) R and call the quotient we get Q(x). We can say that P(x)/(x-a) = Q(x) + R/(x-a) If we multiply both sides by (x-a), we get: P(x) = Q(x)(x-a) + R If x = a, (x-a) = 0, so we have P(a) = R If R = 0, then, P(a) has a zero at a.

What’s the Point? You might be asking yourself why we’d ever want to do long division to find the value of a function when we can just plug in our x value. In most cases, long division is time consuming, it’s true, but the remainder theorem also works with synthetic division. Using synthetic division can make this technique faster than plugging in the value.

Example Does P(x) = x 5 +4x 4 -21x 3 -76x 2 +68x+240 have a zero at x = 2? We can synthetically divide the polynomial by (x - 2) in order to find the value of P(x) at that point using the remainder theorem.