If both and are factors of , show that .

Understand the Question According to the Factor Theorem, if is a factor of a polynomial , then . For the polynomial : Since is a factor, . Since is a factor, . We can set up two linear equations in terms of and , equate them, and show that . Step 1 · Apply the Factor Theorem for Let the given polynomial be Since is a factor of , by the Factor Theorem, : Step 2 · Apply the Factor Theorem for Since is a factor of , by the Factor Theorem, : Multiplying the entire equation by : Step 3 · Equate the Two Equations From equations and : Rearranging the terms: Hence proved. Answer > Hence proved, . Common Mistakes Sign Error in Root: Using or instead of solving . Clearing Fractions Error: Forgetting to multiply every term by when simplifying , leading to an incorrect coefficient for .

Exploring Algebraic Identities | EOT

Question 10

If both $x - 2$ and $x - \dfrac{1}{2}$ are factors of $px^2 + 5x + r$ , show that $p = r$ .

Solution

Understand the Question

According to the Factor Theorem, if $(x - a)$ is a factor of a polynomial $P(x)$ , then $P(a) = 0$ .
For the polynomial $P(x) = px^2 + 5x + r$ $P (x) = p x^{2} + 5 x + r$ :
- Since $(x - 2)$ is a factor, $P(2) = 0$ .
- Since $\left(x - \dfrac{1}{2}\right)$ is a factor, $P\left(\dfrac{1}{2}\right) = 0$ .
We can set up two linear equations in terms of $p$ and $r$ , equate them, and show that $p = r$ .

Step 1 · Apply the Factor Theorem for $x - 2$

Let the given polynomial be $P(x) = px^2 + 5x + r$

Since $(x - 2)$ is a factor of $P(x)$ , by the Factor Theorem, $P(2) = 0$ :

\begin{aligned} P(2) &= p(2)^2 + 5(2) + r = 0 \\ 4p + 10 + r &= 0 \\ 4p + r &= -10 \quad \dots (1) \end{aligned}

Step 2 · Apply the Factor Theorem for $x - \dfrac{1}{2}$

Since $\left(x - \dfrac{1}{2}\right)$ is a factor of $P(x)$ , by the Factor Theorem, $P\left(\dfrac{1}{2}\right) = 0$ :

\begin{aligned} P\left(\dfrac{1}{2}\right) &= p\left(\dfrac{1}{2}\right)^2 + 5\left(\dfrac{1}{2}\right) + r = 0 \\[0.6em] \dfrac{p}{4} + \dfrac{5}{2} + r &= 0 \end{aligned}

Multiplying the entire equation by $4$ :

\begin{aligned} p + 10 + 4r &= 0 \\ p + 4r &= -10 \quad \dots (2) \end{aligned}

Step 3 · Equate the Two Equations

From equations $(1)$ and $(2)$ : $4p + r = p + 4r$

Rearranging the terms:

\begin{aligned} 4p - p &= 4r - r \\ 3p &= 3r \\ p &= r \end{aligned}

Hence proved.

Answer

Hence proved, $p = r$ .

Common Mistakes

Sign Error in Root: Using $x = -2$ or $x = -\dfrac{1}{2}$ instead of solving $x - a = 0 \implies x = a$ .
Clearing Fractions Error: Forgetting to multiply every term by $4$ when simplifying $\dfrac{p}{4} + \dfrac{5}{2} + r = 0$ , leading to an incorrect coefficient for $r$ .

Question 10

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