Lesson: Addition and Scalar Multiplication of Polynomials

Question 2

1 point

Transcript

Long before you were introduced to Rn and matrices, you were introduced to polynomials, and somewhere along the way, you were taught addition and scalar multiplication of polynomials, although perhaps not using those terms. Now, we’ll formally define them.

If p(x) is the polynomial a0 + a1x + dot dot dot + an(x-to-the-n) and if q(x) is the polynomial b0 + b1x + dot dot dot + bn(x-to-the-n) are both polynomials of degree less than or equal to n, and if t is a scalar (that is, t is in the real numbers), then there are polynomials (p + q)(x) and (tp)(x) of degree less than or equal to n defined as follows. (p + q)(x) = (a0 + b0) + (a1 + b1)x + dot dot dot + (an + bn)(x-to-the-n) and (tp)(x) = t(a0) + t(a1)x + dot dot dot + t(an)(x-to-the-n).

For example, the scalar product 3(2 + 5x – 4(x-squared)) = 3(2) + 3(5)x + 3(-4)(x-squared), which equals the polynomial 6 + 15x – 12(x-squared).

Now, when adding polynomials, I find it best to line up the corresponding entries in columns. So for example, to add the polynomial 1 + 2x + 6(x-squared) + 5(x-cubed) to the polynomial 3 – x – 9(x-squared) + 3(x-cubed), I would write it like this. And then looking at the first column, we see that we get 1 + 3 = 4. In our second column, we see that +2x – x is a +x. In our third column, we see that +6(x-squared) – 9(x-squared) = -3(x-squared). And in our fourth column, we get that +5(x-cubed) + 3(x-cubed) = +8(x-cubed), so our sum is 4 + x – 3(x-squared) + 8(x-cubed).

This method is particularly useful if some of the coefficients are 0. So for example, if we want to add the polynomial 3 – x-squared + 8(x-to-the-fourth) to the polynomial 2 + 8x – 4(x-squared) + 7(x-cubed) – 5(x-to-the-fifth), we line up the terms as seen here, and then we just add in the columns. So 3 + 2 = 5, the 0 + 8 = 8 for our x term, our -1 + a -4 gives us a -5 for our x-squared term, a 0 + a 7 gives us a +7 for our x-cubed term, 8 + 0 gives us a +8 for our x-to-the-fourth term, and 0 – 5 gives us a -5 for our x-to-the-fifth term. So the sum of these polynomials is 5 + 8x – 5(x-squared) + 7(x-cubed) + 8(x-to-the-fourth) – 5(x-to-the-fifth).

When scalar multiplication and addition are combined, I distribute the scalar first, and then line it up in columns to add. So if we were looking at the linear combination 6 times the polynomial (1 – 3x – 5(x-squared)) + -2 times the polynomial (9 – x-squared), this is the same as the sum of the polynomials 6 – 18x – 30(x-squared) and -18 + 2(x-squared), which we line up in columns like this, and we see that our first column gives us a -12, our second column gives us a -18x, and our third column gives us a -28(x-squared).

So why are we talking about polynomials now? Well, because addition and scalar multiplication of polynomials satisfy the same set of useful properties that we got for both Rn and matrices.

Theorem 4.1.1: Let p(x), q(x), and r(x) be polynomials of degree at most n, and let s and t be in R. Then we have the following properties.

  1. p(x) + q(x) is a polynomial of degree at most n, so the set of polynomials of degree at most n is closed under addition.
  2. The quantity (p(x) + q(x)) + r(x) = p(x) + the quantity (q(x) + r(x)), so addition is associative.
  3. The 0 polynomial, equal to 0 + 0x + dot dot dot + 0(x-to-the-n) satisfies the fact that p(x) + 0 = p(x) = 0 + p(x) for any polynomial p(x). This is called the additive identity property.
  4. For each polynomial p(x), there exists an additive inverse, denoted as a (-p)(x), which has the property that p(x) + (-p)(x) = 0. In particular, (-p)(x) equals the scalar product (-1)(p(x)). This is the additive inverse property.
  5. p(x) + q(x) = q(x) + p(x), so addition is commutative.
  6. s(p(x)) is a polynomial of degree at most n, so our set of polynomials of degree at most n is closed under scalar multiplication.
  7. s(t(p(x))) = (st)(p(x)), so scalar multiplication is associative.
  8. (s + t)(p(x)) = s(p(x)) + t(p(x))—scalar addition is distributive.
  9. s(p(x) + q(x)) = s(p(x)) + s(q(x))—the scalar multiplication is distributive. And,
  10. 1(p(x)) = p(x)—a scalar multiplicative identity.

Now, the proof of this theorem follows easily from the definitions of addition and scalar multiplication. We will not concern ourselves with it just now, but we will take another look at this theorem soon.

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