Question

Finding zeroes and determining multiplicity help please?

For the polynomial function f(x) = (2 - x) * (x + 1) * (x - 3) * (x + 5), find the zeroes. Then determine the multiplicity at each zero and state whether the graph displays the behavior of a touch or a cross at each intercept.

Answer 1

(A) x = -5, cross; x = -1, cross; x = 2, cross; x = 3, cross
(B) x = -5, cross; x = -1, touch; x = 2, touch; x = 3, touch
(C) x = -5, cross; x = -1, cross; x = 2, touch; x = 3, touch
(D) x = -5, cross; x = -1, cross; x = 2, cross; x = 3, touch

Answer 2

i'm not sure but i think A is correct if i understand the question correctly

Answer 3

Hint:
1. make a list of the roots
2. sort them out in ascending order
3. draw a sketch, and compare with the answers.

Answer 4

when it 'touches', the multiplicity is even. when it 'crosses' the multiplicity is odd

Answer 5

@mathmate im still having trouble

Answer 6

@JoeJoldin
in which step?

Answer 7

are my roots 2, 1, -3, and 5?

Answer 8

Close, but not quite.
When the polynomial is, say
\(f(x)=(x-2)(x+3)\)
the roots are such that each factor equal zero, i.e. (x-2)=0, or (x+3)=0, which makes the zeroes x=2 or x-3.
Ok so far?

Answer 9

@JoeJoldin ok so far?

Answer 10

Can you now list the zeroes?

Answer 11

@mathmate 5, 1, -2, -3?

Answer 12

Recall that with the factor (x-2), the zero is obtained by equating to zero (x-2)=0, which, when solved for x, gives x=\( +\)2, the zero for the (x+3) factor is x=-3.

So the sign of the zeros that you obtained need to be reversed.

You need to sort them in ascending order and proceed to step 3.
Since this is a quartic (highest degree is 4, with the leading coefficient negative, the graph looks like:


where the 4 intersections with the x-axis corresponding to the roots.
Since there is no repeated roots (such as (x-5)(x-5)=(x-5)^2, it won't be "touching" the x-axis.