Express f(x) in the form f(x)=(x-k) q(x)+r for the given value of k. f(x)=x^3+4 x^2+5 x+2 ; k=-2

Name: Problem 21, Chapter 3: Precalculus Student Solutions Manual 5th
Uploaded: 2020-08-05T13:35:18-08:00
Duration: 3 min 28 s
Channel: Cullen Miller
Description: Express f(x) in the form f(x)=(x-k) q(x)+r for the given value of k. f(x)=x^3+4 x^2+5 x+2 ; k=-2

step 1 Hello, hope you're doing well. So we've got our f of x function here and our k value here. And w

Express f(x) in the form f(x)=(x-k) q(x)+r for the given...

Key Concepts

Polynomial Division

Polynomial Division is the process of dividing one polynomial by another, typically a binomial, to express the original polynomial as a product of the divisor and a quotient, plus a remainder. This division is governed by the Division Algorithm for polynomials, which states that for any two polynomials f(x) and d(x) (with d(x) nonzero), there exist unique polynomials q(x) (the quotient) and r(x) (the remainder) such that f(x) = d(x)q(x) + r(x), where the degree of r(x) is less than the degree of d(x).

Remainder Theorem

The Remainder Theorem is a special case of polynomial division which asserts that if a polynomial f(x) is divided by a linear divisor of the form (x - a), then the remainder of this division is simply f(a). This theorem provides a quick method to compute the remainder without performing the full division, and it is essential in understanding polynomial roots and factorization.

Synthetic Division

Synthetic Division is a streamlined technique for dividing a polynomial by a binomial of the form (x - a). It simplifies the division process through a series of arithmetic operations that focus on the coefficients of the polynomial, making it faster and less prone to error compared to long division. Synthetic Division is particularly useful when applying the Remainder Theorem or when factoring polynomials.

Transcript

00:01 Hello, hope you're doing well.

00:02 So we've got our f of x function here and our k value here.

00:05 And we need to get this f of x function in the form x minus k times q of x plus r.

00:12 So in order to do that, we are first going to take this f of x function.

00:20 We are essentially going to use synthetic division to divide this by x minus k.

00:27 K in this case is minus 2, so it's going to be x minus minus 2 or x plus 2.

00:32 So we're going to divide this by x plus 2.

00:36 So we're going to use synthetic division to do this.

00:38 So we're going to take our numerator and write down all of its coefficients.

00:43 We've got 1, 4, 5, 2.

00:51 And then again, remember from before, our k values equal to minus 2.

00:55 So we just put this on the outside.

00:57 So using synthetic division to solve this, we can carry down this 1.

01:01 We have 1 times minus 2 gives us minus 2, 4 minus 2 gives us 2.

01:06 2 times minus 2 gives us minus 4, 5 minus 4 gives us 1, 1 times minus 2 gives us minus 2, and 2 minus 2 gives us 0.

01:15 So this right here is our remainder term.

01:17 This is our x -to -the -zero term, or our constant.

01:20 This is our x -term, and this is our x -squared term.

01:23 So it means that the result of this synthetic division is going to be 1x squared or x -squared plus 2x, and then our constant is just 1.

01:33 And then our remainder is 0, so we don't have to worry about our remainder...

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