By the Remainder Theorem,find the remainder w | vedclass

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(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(ax - b)$,the remainder is $p\left(\frac{b}{a}\right)$.Given $g(x) = 2x -

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$\frac{3}{2}$

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(A) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(ax - b)$,the remainder is $p\left(\frac{b}{a}\right)$.Given $g(x) = 2x - 1$,we set $g(x) = 0$ to find the value of $x$:$2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = \frac{1}{2}$.Now,substitute $x = \frac{1}{2}$ into $p(x)$:$p\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 14\left(\frac{1}{2}\right) - 3$.$p\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) + 7 - 3$.$p\left(\frac{1}{2}\right) = \frac{1}{2} - 3 + 7 - 3$.$p\left(\frac{1}{2}\right) = \frac{1}{2} + 1 = \frac{3}{2}$.Thus,the remainder is $\frac{3}{2}$.

By the Remainder Theorem,find the remainder when $p(x)$ is divided by $g(x)$,where $p(x) = 4x^3 - 12x^2 + 14x - 3$ and $g(x) = 2x - 1$.

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