Question

1. the graph w has 3 zeros at 0, 5, -6/7. select all the equations that could represent

a. $-2x(x + 5)\left(x - \frac{6}{7}\
ight)$
b. $-3x(x - 5)\left(x + \frac{6}{7}\
ight)$
c. $x(x - 5)(6x + 7)$
d. $x(x - 5)(7x + 6)$
e. $-x(x - 5)(7x + 6)^2$
f. $x^2(x - 5)^2(7x + 6)$

Explanation:

Step1: Recall Zero Product Property

If a polynomial has a zero at \( x = a \), then \( (x - a) \) is a factor. For zero at \( x = 0 \), factor is \( x \); at \( x = 5 \), factor is \( (x - 5) \); at \( x=-\frac{6}{7} \), solve \( 7x + 6 = 0 \Rightarrow x = -\frac{6}{7} \), so factor is \( (7x + 6) \). The leading coefficient can be any non - zero real number, and we can also have repeated factors (but the zeros given are 0, 5, \( -\frac{6}{7} \), so repeated factors of these zeros are allowed, but new zeros are not).

Step2: Analyze Option a

Factor for \( x = -\frac{6}{7} \) should be \( (x+\frac{6}{7})=\frac{7x + 6}{7} \), but the factor here is \( (x-\frac{6}{7}) \), which would correspond to zero at \( x=\frac{6}{7} \), not \( -\frac{6}{7} \). So option a is incorrect.

Step3: Analyze Option b

Factor for \( x = -\frac{6}{7} \): \( (x+\frac{6}{7})=\frac{7x + 6}{7} \), the factor here is \( (x+\frac{6}{7}) \), and we have factors \( x \) (zero at 0), \( (x - 5) \) (zero at 5), \( (x+\frac{6}{7}) \) (zero at \( -\frac{6}{7} \)). The leading coefficient is - 3 (non - zero), so this polynomial has zeros at 0, 5, \( -\frac{6}{7} \). So option b is correct.

Step4: Analyze Option c

Factor for \( x=-\frac{6}{7} \): Solve \( 6x+7 = 0\Rightarrow x=-\frac{7}{6}
eq-\frac{6}{7} \). So this factor does not correspond to zero at \( -\frac{6}{7} \). Option c is incorrect.

Step5: Analyze Option d

We have factors \( x \) (zero at 0), \( (x - 5) \) (zero at 5), and \( (7x + 6) \) (zero at \( x=-\frac{6}{7} \)). The leading coefficient is 1 (non - zero), so this polynomial has zeros at 0, 5, \( -\frac{6}{7} \). So option d is correct.

Step6: Analyze Option e

We have factors \( x \) (zero at 0), \( (x - 5) \) (zero at 5), and \( (7x + 6)^2 \) (zero at \( x = -\frac{6}{7} \), with multiplicity 2). The leading coefficient is - 1 (non - zero), so this polynomial has zeros at 0, 5, \( -\frac{6}{7} \) (with multiplicity 2 for \( -\frac{6}{7} \)). So option e is correct.

Step7: Analyze Option f

We have factors \( x^2 \) (zero at 0 with multiplicity 2), \( (x - 5)^2 \) (zero at 5 with multiplicity 2), and \( (7x + 6) \) (zero at \( -\frac{6}{7} \)). The polynomial has zeros at 0, 5, \( -\frac{6}{7} \) (with multiplicities 2, 2, 1 respectively). So option f is correct.

Answer:

b. \( -3x(x - 5)(x+\frac{6}{7}) \), d. \( x(x - 5)(7x + 6) \), e. \( -x(x - 5)(7x + 6)^2 \), f. \( x^2(x - 5)^2(7x + 6) \)