Question

Find the zeroes of the polynomial $p(x) = 3x^2 - 4x - 4$. Hence, write a polynomial whose each of the zeroes is 2 more than the zeroes of $p(x)$.

Hint:

To shift roots, replace $x$ with $x - k$ in the original roots.

Answer 1

Given:
Original polynomial: \( p(x) = 3x^2 - 4x - 4 \)

The objective is to determine a new polynomial whose roots exceed the roots of the given polynomial by 2.

Step 1: Determine the roots of the original polynomial using the quadratic formula
\[x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}= \frac{4 \pm \sqrt{16 + 48}}{6}= \frac{4 \pm \sqrt{64}}{6}= \frac{4 \pm 8}{6}\]
Thus, the roots are:
\[x = \frac{4 + 8}{6} = \frac{12}{6} = 2,\quadx = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}\]

Step 2: Increase each root by 2
The new polynomial must have roots that are 2 greater than the original roots.
Therefore, the new roots are:
- \( 2 + 2 = 4 \)
- \( -\frac{2}{3} + 2 = \frac{4}{3} \)

Step 3: Construct the polynomial from the new roots
Given the roots \( x = 4 \) and \( x = \frac{4}{3} \), the polynomial is:
\[(x - 4)\left(x - \frac{4}{3}\right)\]
Expanding the factors:
\[= x^2 - \left(4 + \frac{4}{3}\right)x + 4 \cdot \frac{4}{3}= x^2 - \frac{16}{3}x + \frac{16}{3}\]
Step 4: Eliminate the fraction by multiplying the entire expression by 3
\[3 \left( x^2 - \frac{16}{3}x + \frac{16}{3} \right)= 3x^2 - 16x + 16\]
Final Answer:
The required polynomial is: \( \boxed{3x^2 - 16x + 16} \)