If both (x-2) and (x-½) are factors of px²+5x+r, show that p=r

Given that both \((x - 2)\) and \((x - \frac{1}{2})\) are factors of the quadratic expression \(px^2 + 5x + r\), we need to show that \(p = r\).

Since \((x - 2)\) and \((x - \frac{1}{2})\) are factors, the quadratic can be expressed as:
\[
px^2 + 5x + r = p(x - 2)(x - \frac{1}{2})
\]

Expanding the right-hand side:

\[
p(x - 2)(x - \frac{1}{2}) = p\left(x^2 - \frac{1}{2}x - 2x + 1\right)
\]

Simplify the expression inside the parentheses:

\[
p\left(x^2 - \frac{5}{2}x + 1\right)
\]

Distribute \(p\):

\[
px^2 - \frac{5}{2}px + p
\]

Now, compare the expanded form \(px^2 - \frac{5}{2}px + p\) with the original quadratic expression \(px^2 + 5x + r\).

By comparing coefficients:

1. The coefficient of \(x^2\) is \(p\), which matches on both sides.
2. The coefficient of \(x\) gives us:
   \[
   -\frac{5}{2}p = 5
   \]

3. The constant term comparison gives:
   \[
   p = r
   \]

Let's solve for \(p\) from the second equation:
\[
-\frac{5}{2}p = 5
\]
\[
p = -\frac{5}{2} \times \frac{1}{5} = -2
\]

Since \(p = r\), we also have \(r = -2\).

Thus, we have shown that \(p = r\).

Would you like more details or have any questions? Here are five related questions to explore:

1. What is the significance of comparing coefficients in polynomial equations?
2. How can the factor theorem be used to find unknown coefficients?
3. Why is it important to check that both sides of an equation are balanced?
4. Can this method be used for polynomials of higher degrees?
5. What are some other methods to factor quadratic equations?

**Tip:** Always ensure you compare coefficients correctly when expanding and matching polynomial equations.

Explore how to prove that p equals r when both (x - 2) and (x - 1/2) are factors of px^2 + 5x + r. This detailed solution covers polynomial equations, quadratic expansion, and the factor theorem. Suitable for advanced high school students and college-level algebra courses.

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