Find the root of the factor (2x−1).

Question

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Here are the solutions to the questions:

Find the root of the factor $(2x-1)$ .
Set $2x-1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}$ .
Substitute $x = \frac{1}{2}$ into the polynomial $f(x)$ .
$f\left(\frac{1}{2}\right) = 12\left(\frac{1}{2}\right)^3 - 28\left(\frac{1}{2}\right)^2 - 3\left(\frac{1}{2}\right) + 7$
Calculate the value of $f\left(\frac{1}{2}\right)$ .
$f\left(\frac{1}{2}\right) = 12\left(\frac{1}{8}\right) - 28\left(\frac{1}{4}\right) - \frac{3}{2} + 7$ $f\left(\frac{1}{2}\right) = \frac{12}{8} - \frac{28}{4} - \frac{3}{2} + 7$ $f\left(\frac{1}{2}\right) = \frac{3}{2} - 7 - \frac{3}{2} + 7$ $f\left(\frac{1}{2}\right) = 0$
Conclude using the Factor Theorem.
Since $f\left(\frac{1}{2}\right) = 0$ , by the Factor Theorem, $(2x-1)$ is a factor of $f(x)$ .

2.2 The expression $x^3 - ax^2 + ax + b$ is divisible by $(x+1)$ , $(x-3)$ and $(x+m)$ .

Let $P(x) = x^3 - ax^2 + ax + b$ .

2.2.1 Determine the values of $a$ and $b$ .
Use the Factor Theorem for $(x+1)$ .
Since $P(x)$ is divisible by $(x+1)$ , $P(-1) = 0$ .
$P(-1) = (-1)^3 - a(-1)^2 + a(-1) + b = 0$ $-1 - a - a + b = 0$ $-1 - 2a + b = 0 \implies b = 2a + 1 \quad \text{(Equation 1)}$
Use the Factor Theorem for $(x-3)$ .
Since $P(x)$ is divisible by $(

Find the root of the factor (2x-1).