Polynomial Multiplication!

Alright! Let’s dive into the fascinating world of Polynomial Multiplication! ✨ We’ve already covered addition and subtraction, which are about combining “like terms”. Multiplication introduces a new twist, but it’s just as logical and easy to grasp. Get ready for some algebra fun! 🚀

What is a Polynomial Again? (A quick mental stretch! 🧠)

Just to quickly recap from our previous conversation, a polynomial is a mathematical expression that’s essentially a sum of several mathematical terms. Each of these terms can be a number, a variable, or a product of several variables, like 3x or x²y. From a mathematician’s perspective, the operations allowed are addition, subtraction, multiplication, and variables must have “natural exponents” (non-negative integers like 0, 1, 2, etc.).

Algebra of Polynomials: Multiplication ✖️

When multiplying polynomials, the core idea is to make sure every term in one polynomial is multiplied by every term in the other polynomial. Then, you combine any “like terms” that result from these multiplications. It’s like a grand matchmaking event where every term gets a partner from the other side! 🥰

Let’s break it down into simpler steps:

1. Multiplying a Polynomial by a Monomial (Single Term) 💡

This is the most straightforward case. You simply distribute the monomial to each term in the polynomial. Remember the Law of Exponents here: when you multiply variables with the same base, you add their exponents (e.g., $x^2 \times x^3 = x^{2+3} = x^5$).

Example: Multiply 2x³ by (x² + x + 1).

2x³ * (x² + x + 1) $= (2x³ \times x²) + (2x³ \times x) + (2x³ \times 1) $= (2 \times x^{3+2}) + (2 \times x^{3+1}) + (2 \times x³)$ $= 2x⁵ + 2x⁴ + 2x³ ✨

2. Multiplying Two Polynomials (General Case) 🧑‍🏫

This extends the idea of distributing. If you have two polynomials, you can break down one polynomial into its individual monomial terms, and then multiply each of those monomials by the entire second polynomial. After all the multiplications are done, you add the resulting like terms.

This is the generalization of the “FOIL” method (First, Outer, Inner, Last) that you might have used for multiplying two binomials (polynomials with two terms). FOIL essentially applies this distributive principle for two terms.

Example: Multiply (x + 1) by (x² + x + 1).

Think of (x + 1) as two separate monomials: x and 1. Then, distribute each of these to (x² + x + 1):

x * (x² + x + 1) = x³ + x² + x + 1 * (x² + x + 1) = x² + x + 1

Now, add the results and combine any like terms: (x³ + x² + x) + (x² + x + 1) $= x³ + (x² + x²) + (x + x) + 1 $= x³ + 2x² + 2x + 1 🥳

General Formula & Resultant Degree 📝

For a more formal approach, if you have two polynomials, p(x) of degree n and q(x) of degree m: p(x) = a₀ + a₁x + a₂x² + ... + aₙxⁿ q(x) = b₀ + b₁x + b₂x² + ... + bₘxᵐ

Their product p(x)q(x) will be a polynomial whose degree is the sum of their individual degrees, i.e., n + m.

The coefficient of any term x^k in the product p(x)q(x) is found by summing all products a_j * b_{k-j} where j ranges from 0 to k (or up to the minimum of n or m, depending on the specific coefficients available).

So, p(x)q(x) = ∑_{k=0}^{n+m} (∑_{j=0}^{k} a_j b_{k-j})x^k.

This formula just formalizes the “multiply every term by every term” rule. For example, to find the coefficient of x² (where k=2), you’d sum a₀b₂ + a₁b₁ + a₂b₀.

Key Insight: The product of two polynomials will always result in another polynomial.

Practice Questions! 🌟

Question 1: Multiply the following: P(x) = 5x² Q(x) = 3x³ - 2x + 7

Question 2: Multiply the following: A(x) = (x + 3) B(x) = (x - 5)

Question 3: Multiply the following: C(x) = (2x² + x - 1) D(x) = (x + 4)

Question 4: Given E(x) = x³ + 2x - 1 and F(x) = x² - x + 3, find the coefficient of x³ in E(x)F(x).

Solutions to Practice Questions ✅

Solution 1: Multiply P(x) = 5x² and Q(x) = 3x³ - 2x + 7

Here, we multiply the monomial 5x² by each term in Q(x): P(x)Q(x) = 5x² * (3x³ - 2x + 7) = (5x² \times 3x³) + (5x² \times -2x) + (5x² \times 7) = (5 \times 3 \times x^{2+3}) + (5 \times -2 \times x^{2+1}) + (5 \times 7 \times x²) = 15x⁵ - 10x³ + 35x² 🎯

Solution 2: Multiply A(x) = (x + 3) and B(x) = (x - 5)

Multiply each term of A(x) by each term of B(x): A(x)B(x) = x * (x - 5) + 3 * (x - 5) = (x \times x) + (x \times -5) + (3 \times x) + (3 \times -5) = x² - 5x + 3x - 15 Now, combine like terms (-5x and 3x): = x² + (-5 + 3)x - 15 = x² - 2x - 15 🥳

Solution 3: Multiply C(x) = (2x² + x - 1) and D(x) = (x + 4)

Multiply each term of C(x) by each term of D(x): C(x)D(x) = 2x² * (x + 4) + x * (x + 4) - 1 * (x + 4) = (2x² \times x) + (2x² \times 4) + (x \times x) + (x \times 4) + (-1 \times x) + (-1 \times 4) = 2x³ + 8x² + x² + 4x - x - 4 Now, combine like terms (8x² and x², and 4x and -x): = 2x³ + (8 + 1)x² + (4 - 1)x - 4 = 2x³ + 9x² + 3x - 4 ✨

Solution 4: Given E(x) = x³ + 2x - 1 and F(x) = x² - x + 3, find the coefficient of x³ in E(x)F(x).

To find the coefficient of x³ (where k=3), we need to identify pairs of terms from E(x) and F(x) whose product results in an x³ term. Using the general formula a_j b_{k-j} x^k:

For x³ (k=3), we look for a_j * b_{3-j} combinations:

• j=0: a₀ * b₃ (constant from E(x) * x³ term from F(x))
  • In E(x), a₀ = -1. In F(x), there is no x³ term (coefficient b₃ = 0). So, (-1) * 0 = 0.
• j=1: a₁ * b₂ (x term from E(x) * x² term from F(x))
  • In E(x), a₁ = 2 (from 2x). In F(x), b₂ = 1 (from x²). So, (2) * (1) = 2.
• j=2: a₂ * b₁ (x² term from E(x) * x term from F(x))
  • In E(x), there is no x² term (coefficient a₂ = 0). In F(x), b₁ = -1 (from -x). So, (0) * (-1) = 0.
• j=3: a₃ * b₀ (x³ term from E(x) * constant from F(x))
  • In E(x), a₃ = 1 (from x³). In F(x), b₀ = 3. So, (1) * (3) = 3.

Now, sum these products to get the total coefficient for x³: 0 + 2 + 0 + 3 = 5

Therefore, the coefficient of x³ in E(x)F(x) is 5. 🌟