1) Since x = -4 and x = -6 are zeros, then, by definition, (x + 4) and (x + 6) are factors of the polynomial. Another way in understanding this is that when x = -4, there would be no remainders.
Thus, we can simply write it as: P(x) = (x + 4)(x + 6) = x² + 10x + 24 (ie C)
2) Using the discriminant, we can find whether a quadratic has two, one, or no roots at all. Symbolised as delta (Δ), the discriminant formula goes as follows:
Δ = b² - 4ac Thus, if we move the 3 to the other side, we get: Δ = (-5)² - 4(3)(-10) > 0 Since Δ > 0, then there lies two solutions.
To find rationality or irrationality, we need to check whether or not the square root of the discriminant is an integer or not. If the square root of the discriminant is not an integer, we have irrational roots.
Thus, we've worked out that the discriminant is 145. But 145 is not a perfect square, so when we take the square root of it, we get irrational roots. Hence, we have two irrational roots.
3) Since we have a fraction, we know that the denominator cannot be zero. Thus, to find places where undefined x-values occur, we need to let the denominator equal to zero:
x² - 16 = 0 x² = 16 x = 4 and -4
These are the restricted values for x, since x cannot equal to 4, or -4. Hence, your answer is x = 4, -4
4) We can expand and move everything to the LHS. 2(x + 2)² - 32 = 0 2(x² + 4x + 4) - 32 = 0 2x² + 8x + 8 - 32 = 0 2x² + 8x - 24 = 0
