The zeroes of the quadratic polynomial \(x^2 + 99x + 127\) are
both positive
both negative
one positive and one negative
both equal
Step 1: Recall the relations between coefficients and zeroes.
For quadratic polynomial \(ax^2 + bx + c\):
Sum of zeroes = \(-\dfrac{b}{a}\),
Product of zeroes = \(\dfrac{c}{a}\).
Step 2: Apply to the given polynomial.
Here, \(a = 1\), \(b = 99\), \(c = 127\).
So,
Sum = \(-\dfrac{99}{1} = -99\).
Product = \(\dfrac{127}{1} = 127\).
Step 3: Interpret the signs.
Since product \(127 > 0\), both zeroes must have the same sign.
Since sum \(-99 < 0\), the common sign must be negative.
Conclusion: Both zeroes are negative.