Question

Find the polynomial f(x) that has the roots of -3, 5 of multiplicity 2. Explain how you would verify the zeros of f(x)

Answer 1

@anas2000

Answer 2

turn roots ==> factors, and multiply

x = -2 ==> x + 2
x = 5 ==> x - 5
multiplicity 2 ==> (x - 5)^2

f(x) = a(x + 2)(x - 5)^2
f(x) = a(x + 2)(x^2 - 10x + 25)
f(x) = a(x^3 - 8x^2 + 5x + 50)

you can test this synthetically or with the factor theorem

f(-2) = a(-8 - 8(4) + 5(-2) + 50) = a(0) = 0... check, f = -2 works
f(5) = a(125 - 8(25) + 5(5) + 50) = a(0) = 0 check, f = 5 works

will have to divide to check the root of multiplicity two

5 | 1 .... -8 ..... 5 ..... 50
............5...... -15. ...-50
...--------------------------------
....1 .....-3 .....-10..... 0 <== remainder is 0, confirming x - 5 is a factor, and x = 5 is a root

quotient: x^2 - 3x - 10 = (x - 5)(x + 2), confirming the other two zeros are x = 5 (again), and x = -2

Answer 3

The factors corresponding to given roots are (x + 2) and (x - 5)². Hence we start with
f(x) = k(x + 2)(x - 5)² [family of functions with the given roots]
For a particular member of the family k ∈ ℝ. Suppose k = 1, then
f(x) = (x + 2)(x - 5)² is one example of the polynomial which has roots -2, 5 of multiplicity 2)Zeros of a function are the same as the roots of an equation (the x- intercepts). How does one find the x-intercepts? By letting y = 0 and solving for x. If an equation is in factored form, equate each variable factor to 0 and solve for x. Why? Because factors are terms you multiply to get the product, and a product equals 0 when any factor is 0.

Answer 4

omg.......wow!!!!!!!!!!

Answer 5

hope this helped

Answer 6

Hey @PassionGirl93 @anas2000
Where'd the (x + 2) come from? There is no instruction that says there must be a root of -2

Answer 7

I'm wondering the same

Answer 8

They had the right idea, though... The gist of it is, if you want a polynomial with a root r, just include
(x - r) among the factors...
If you want it to be of multiplicity k
Then include
(x - r)^k
among the factors.

Answer 9

ok thank you:))