Questions: Consider the following 3 expressions. To be the same, what would P, Q, and R have to be? P x^2 + 6 x + 1 x^2 + x(x + Q) + 1 3 x^2 - 5 - (x^2 - 6 x + R/2) P= Q= R=

Solution Steps

To determine the values of \( P \), \( Q \), and \( R \) that make the given expressions equivalent, we need to equate the coefficients of the corresponding terms in each expression. This involves expanding and simplifying the expressions and then comparing the coefficients of \( x^2 \), \( x \), and the constant terms.

We start with the first expression: \[ P x^2 + 6x + 1 \]

Next, we look at the second expression: \[ x^2 + x(x + Q) + 1 \] Simplify the second expression: \[ x^2 + x^2 + Qx + 1 \] \[ 2x^2 + Qx + 1 \]

Now, consider the third expression: \[ 3x^2 - 5 - \left(x^2 - 6x + \frac{R}{2}\right) \] Simplify the third expression: \[ 3x^2 - 5 - x^2 + 6x - \frac{R}{2} \] \[ 2x^2 + 6x - 5 - \frac{R}{2} \]

We need to compare the simplified forms of the three expressions to find \( P \), \( Q \), and \( R \).

1. From the first and second expressions: \[ P x^2 + 6x + 1 = 2x^2 + Qx + 1 \] By comparing coefficients: \[ P = 2 \] \[ 6 = Q \]

2. From the first and third expressions: \[ P x^2 + 6x + 1 = 2x^2 + 6x - 5 - \frac{R}{2} \] By comparing coefficients: \[ P = 2 \] \[ 1 = -5 - \frac{R}{2} \] Solve for \( R \): \[ 1 + 5 = -\frac{R}{2} \] \[ 6 = -\frac{R}{2} \] \[ R = -12 \]

Final Answer