Find the value of k for which the expression \(x^3 - 3x^2 - 7x - k\) will be divisible by \((x - 5)\).
15
By the remainder theorem, a polynomial \(f(x)\) is divisible by \((x - 5)\) exactly when \(f(5) = 0\).
Here \(f(x) = x^3 - 3x^2 - 7x - k\), so substitute x = 5.
Compute each term: \(5^3 = 125\), \(3 \times 5^2 = 75\), and \(7 \times 5 = 35\).
So \(f(5) = 125 - 75 - 35 - k = 15 - k\).
Set this equal to zero: \(15 - k = 0\), giving \(k = 15\).
Hence, the required value of k is 15.
What is the remainder of function $4a^3 - 12a^2 + 14a - 3$ when it is divided by \(\frac{a-1}{2}\)?
If $(x + 1)$ and $(x + 2)$ are factors of $ax^3 + 3x^2 + bx$ then the values of $a$ and $b$ are:
Solve the following.
Subtract $\frac{9}{2} + \frac{x}{2} + \frac{3}{5}x^2 + \frac{7}{4}x^3$ from $\frac{7}{2} - \frac{x}{3} - \frac{1}{5}x^2$.