Given that p(x) leaves a remainder of R when divided by (x-1), we have p(1) = R.

Hey Ria, good to see you again. a) Find the values of $R$ and $k$. Step 1: Use the Remainder Theorem for division by $(x-1)$. Given that $p(x)$ leaves a remainder of $R$ when divided by $(x-1)$, we have $p(1) = R$. Substitute $x=1$ into $p(x) = x^3 + kx^2 - x + 12$: $$p(1) = (1)^3 + k(1)^2 - (1) + 12$$ $$R = 1 + k - 1 + 12$$ $$R = k + 12 \quad (*)$$ Step 2: Use the Remainder Theorem for division by $(x-4)$. Given that $p(x)$ leaves a remainder of $8R$ when divided by $(x-4)$, we have $p(4) = 8R$. Substitute $x=4$ into $p(x) = x^3 + kx^2 - x + 12$: $$p(4) = (4)^3 + k(4)^2 - (4) + 12$$ $$8R = 64 + 16k - 4 + 12$$ $$8R = 16k + 72 \quad (**)$$ Step 3: Solve the system of equations for $R$ and $k$. Substitute equation $()$ into equation $(*)$: $$8(k + 12) = 16k + 72$$ $$8k + 96 = 16k + 72$$ $$96 - 72 = 16k - 8k$$ $$24 = 8k$$ $$k = \frac{24}{8}$$ $$k = 3$$ Now substitute $k=3$ back into equation $(*)$ to find $R$: $$R = 3 + 12$$ $$R = 15$$ The values are $\boxed{R=15, k=3}$. b) Show that $(x+4)$ is a factor of $p(x)$. Step 1: Update $p(x)$ with the value of $k$. Using $k=3$, the polynomial is $p(x) = x^3 + 3x^2 - x + 12$. Step 2: Apply the Factor Theorem. For $(x+4)$ to be a factor of $p(x)$, we must show that $p(-4) = 0$. Substitute $x=-4$ into $p(x)$: $$p(-4) = (-4)^3 + 3(-4)^2 - (-4) + 12$$ $$p(-4) = -64 + 3(16) + 4 + 12$$ $$p(-4) = -64 + 48 + 4 + 12$$ $$p(-4) = -64 + 64$$ $$p(-4) = 0$$ Since $p(-4) = 0$, by the Factor Theorem, $(x+4)$ is a factor of $p(x)$. Send me the next one 📸