# Exploring the Relationship Between Algebraic Expressions

## If (px + 5)(qx + 3) = 8x^2 + rx + 15 for all values of x

When we are given a quadratic equation in the form of (px + 5)(qx + 3) = 8x^2 + rx + 15 for all values of x, we can use this information to find the relationship between the coefficients p, q, and r. In this case, we are also given that p + q = 6.

To begin solving this equation, we need to expand the left side using the distributive property. This will give us two terms: one with x^2, one with x, and a constant term. After expanding and simplifying, we have:

(px + 5)(qx + 3) = p*q*x^2 + 3px + 5qx + 15

= pqx^2 + 3px + 5qx + 15

Now we can compare this expression to the given equation 8x^2 + rx + 15. By equating the coefficients of like terms, we can create two equations:

pq = 8 (coefficient of x^2)

3p + 5q = r (coefficient of x)

Since p + q = 6, we can rewrite this equation as p = 6 - q and substitute it into the equation 3p + 5q = r:

3(6 - q) + 5q = r

18 - 3q + 5q = r

18 + 2q = r

Now, we also know that pq = 8. Substituting p = 6 - q into this equation, we get:

(6 - q)q = 8

6q - q^2 = 8

q^2 - 6q + 8 = 0

By solving this quadratic equation, we find that the two possible values for q are 2 and 4. Substituting these values back into p = 6 - q, we get p = 4 and p = 2 respectively.

## What are the two possible values for r?

Now that we have found the values for p and q, we can substitute them back into the equation 18 + 2q = r, which we derived earlier:

When q = 2:

18 + 2(2) = r

18 + 4 = r

r = 22

When q = 4:

18 + 2(4) = r

18 + 8 = r

r = 26

Therefore, the two possible values for r are 22 and 26.

**i hope this will help you !**