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

We are to find a new polynomial whose roots are each 2 more than the roots of the given polynomial.

Step 1: Use the quadratic formula to find roots of the original polynomial
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6} \]
So, the two roots are:
\[ x = \frac{4 + 8}{6} = \frac{12}{6} = 2,\quad x = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3} \]

Step 2: Increase each root by 2
We are told to find a new polynomial with roots that are 2 more than the original roots.
So, new roots are:
- \( 2 + 2 = 4 \)
- \( -\frac{2}{3} + 2 = \frac{4}{3} \)

Step 3: Form polynomial from new roots
If roots are \( x = 4 \) and \( x = \frac{4}{3} \), then the polynomial is:
\[ (x - 4)\left(x - \frac{4}{3}\right) \]
Multiply 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: Clear the fraction by multiplying 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: \( \boxed{3x^2 - 16x + 16} \)