Problem 62 - Algebra Form and Function

Question

Find the constants \(r, s, p,\) and \(q\) if multiplying out the polynomial \(\left(r x^{5}+2 x^{4}+3\right)\left(2 x^{3}-s x^{2}+p\right)\) gives \(6 x^{8}-11 x^{7}-10 x^{6}-12 x^{5}-8 x^{4}+q x^{3}-15 x^{2}-12 .\)

Step-by-Step Solution

Answer

Question: Find the values of the constants \(r, s, p, q\) in the given polynomials: \((r x^{5}+2 x^{4}+3)\left(2 x^{3}-s x^{2}+p\right) = 6 x^{8} - 11 x^{7} - 10 x^{6} - 12 x^{5} - 8 x^{4} + q x^{3} - 15 x^{2} -12\) Answer: \(r = 3, s = 5, p =- 4, q = -9\)

Step 1: Multiply the given polynomials

We will multiply the given polynomials using the distributive property (also known as FOIL). This means that we will multiply each term in the first polynomial by each term in the second polynomial and then add the products together. So we have \((r x^{5}+2 x^{4}+3)\left(2 x^{3}-s x^{2}+p\right)\):

Step 2: Apply FOIL

We will now apply FOIL to multiply the terms in the given polynomials and then sum them up: \(r x^{5} \cdot 2 x^{3} = 2r x^{8}\) \(r x^{5} \cdot (-s) x^{2} = -rs x^{7}\) \(r x^{5} \cdot p = rp x^{5}\) \(2 x^{4} \cdot 2 x^{3} = 4 x^{7}\) \(2 x^{4} \cdot (-s) x^{2} = -2s x^{6}\) \(2 x^{4} \cdot p = 2p x^{4}\) \(3 \cdot 2 x^{3} = 6 x^{3}\) \(3 \cdot (-s) x^{2} = -3s x^{2}\) \(3 \cdot p = 3p\) Now, sum up all the terms: \(2r x^{8} - rs x^{7} + rp x^{5} + 4 x^{7} - 2s x^{6} + 2p x^{4} + 6 x^{3} - 3s x^{2} + 3p\)

Step 3: Compare coefficients with the given result

Now we compare the corresponding coefficients of each term in our product with the given result: \(6 x^{8} - 11 x^{7} - 10 x^{6} - 12 x^{5} - 8 x^{4} + q x^{3} - 15 x^{2} -12\) This gives us the following system of equations: \(2r = 6\) \(-rs + 4 = -11\) \(rp = -12\) \(-2s = -10\) \(2p = -8\) \(6 - 3s + 3p = q\) \(-3s = -15\) Now, we can solve these equations step by step:

Step 4: Solve the equations

First, let's solve the simplest of the equations: \(2r = 6 \implies r = 3\) \(-2s = -10 \implies s = 5\) \(2p = -8 \implies p = -4\) Now, plug the values of \(r\), \(s\), and \(p\) into the remaining equations. Start with the equation for \(rp\): \(rp = -12 \implies 3 \cdot (-4) = -12\) Now, plug the values into the equation for the \(-rs + 4\) term: \(-rs + 4 = -11 \implies -3 \cdot 5 + 4 = -11\) Finally, find the value of \(q\): \(6 - 3s + 3p = q \implies 6 - 3 \cdot 5 + 3 \cdot (-4) = q \implies -9 = q\) Thus, our constants are: \(r = 3, s = 5, p =- 4, q = -9\)

Key Concepts

Coefficients comparison

In polynomial multiplication problems, comparing coefficients is a crucial step to determine the unknowns in the expression. The process involves matching the coefficients of similar powers of the variable from both sides of the equation. In our scenario, after expanding the given polynomial multiplication, we obtained an expression:

• Expanded: \(2r x^{8} - rs x^{7} + rp x^{5} + 4 x^{7} - 2s x^{6} + 2p x^{4} + 6 x^{3} - 3s x^{2} + 3p\)
• Original: \(6 x^{8} - 11 x^{7} - 10 x^{6} - 12 x^{5} - 8 x^{4} + q x^{3} - 15 x^{2} - 12\)

Comparing coefficients involves aligning terms of the same degree and setting their coefficients equal to each other. For example, for the term involving \(x^8\), we have the equation:

• \(2r = 6\)

By solving these simple linear equations for each power, we find the coefficients, which in turn help us solve for the unknowns \(r\), \(s\), \(p\), and \(q\). This systematic comparison is the foundation for many polynomial problems, where solving a simultaneous set of equations gives the final values.

Distributive property

The distributive property, a fundamental arithmetic principle, is essential for multiplying polynomials. It states that for any integers \(a\), \(b\), and \(c\):

• \(a(b+c) = ab + ac\)

This means each term in one polynomial must be multiplied by each term in the other polynomial. In our problem, we applied this principle to the expression:

• \((r x^{5}+2 x^{4}+3)(2 x^{3}-s x^{2}+p)\)

Each term on the left is distributed over all terms on the right:

• \(r x^{5} \cdot 2 x^{3} = 2r x^{8}\)
• \(r x^{5} \cdot (-s) x^{2} = -rs x^{7}\)
• ... and so on for all combinations.

In polynomial multiplication, using the distributive property ensures that all terms are accounted for, crucial for accurately expanding products and consequently comparing coefficients.

System of equations

Solving a system of equations is required when multiple unknowns must be found simultaneously. In the context of polynomial multiplication, matching coefficients leads naturally to a system of equations.
For our polynomial equation, the expansion and subsequent comparison gave us:

• \(2r = 6\)
• \(-rs + 4 = -11\)
• \(rp = -12\)
• \(-2s = -10\)

To solve this, we start with equations that have a single unknown, such as \(2r = 6\). Solving it gives \(r = 3\). Similarly, \(-2s = -10\) gives \(s = 5\) and \(2p = -8\) results in \(p = -4\). Once you have individual solutions, you can substitute back into other equations to find further unknowns, like \(q\).
This method involves substitution and sometimes elimination techniques, leading to a comprehensive solution where all variables are determined in an orderly manner.