Question

which polynomial function has a real root of multiplicity 2? *
a. $y = 4x^2 - 5x + 5$

b. $y = 4 - 9x^2$

c. $y = 7 - 6x + 3x^2$

d. $y = 4 - 12x + 9x^2$
a
b
c
d

Explanation:

Step1: Recall root multiplicity

A real root of multiplicity 2 means the polynomial can be factored into a square of a linear factor (since multiplicity 2 implies the factor is repeated twice). So we need to factor each polynomial and check the discriminant (for quadratics) or factor form.

Step2: Analyze Option A

For \( y = 4x^2 - 5x + 5 \), calculate discriminant \( D = b^2 - 4ac = (-5)^2 - 4(4)(5) = 25 - 80 = -55 \). Since \( D < 0 \), no real roots. So A is out.

Step3: Analyze Option B

Factor \( y = 4 - 9x^2 = (2 - 3x)(2 + 3x) \). This is a difference of squares, two distinct linear factors, so roots have multiplicity 1. So B is out.

Step4: Analyze Option C

For \( y = 3x^2 - 6x + 7 \) (rewritten), discriminant \( D = (-6)^2 - 4(3)(7) = 36 - 84 = -48 \). \( D < 0 \), no real roots. So C is out.

Step5: Analyze Option D

Factor \( y = 9x^2 - 12x + 4 \). This is a perfect square trinomial: \( (3x - 2)^2 \). So it factors into \( (3x - 2)(3x - 2) \), meaning the root \( x = \frac{2}{3} \) has multiplicity 2.

Answer:

D. \( y = 4 - 12x + 9x^2 \)