Write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros. Write the function in standard form. −2, 3, 6

Question

Please tell me all the steps and show them thouroughly im having a hard time understanding.

Answer 1

Hello Zariah,

Some important conditions:

1. polynomial function
2. least degree
3. the leading coefficient is 1
4. the given zeros -2, 3, 6 are x-intercepts that we have for this function.

-------------------------------------------------------------------------------------------------------------------

I think the most important condition is #4

Because from #4, we know that:

x=-2, y=0 ----------- (x+2)

x= 3, y=0 ----------- (x-3)

x=6, y=0 ----------- (x-6)

Hence,

we can get the polynomial function by multiple above terms

y = (x+2) * (x-3) * (x-6)

it can be tested by plug x=-2, 3, 6 into the equation and see whether y=0

Then,

Let's see what is the leading coefficient of this equation by expanding it.

y = (x+2) * (x-3) * (x-6) = (x²-x-6) * (x-6) = x³-6x²-x²+6x-6x+36 = x³-7x²+36 ----- it is 1

So there is no need to fix anything, the answer is y = (x+2) * (x-3) * (x-6) but we need to write it in standard form, so x³-7x²+36

-------------------------------------------------------------------------------------------

EXTRA：

What if the teacher asks you that the leading coefficient must be 2 instead of 1

the below is the same

--------------------------------------------------------------------------------------------------------------

x=-2, y=0 ----------- (x+2)

x= 3, y=0 ----------- (x-3)

x=6, y=0 ----------- (x-6)

Hence,

we can get the polynomial function by multiple above terms

y = (x+2) * (x-3) * (x-6)

it can be tested by plug x=-2, 3, 6 into the equation and see whether y=0

--------------------------------------------------------------------------------------------------------------

But the leading coefficient "2" needs to be added before (x+2) * (x-3) * (x-6)

So we got : y = 2 * (x+2) * (x-3) * (x-6)

Finally, you can expand this equation and get the standard form in which the polynomial function has given 0s (-2,3,6), has the least degree, and with the leading coefficient of 2

-----------------------------------------------------------------------------------------------------------

Hope it helps!

Answer 2

Suppose a polynomial P(x) has a zero at x = a. That just means P(a) = 0.

If an n-th degree polynomial P(x) is a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + ... + a₁x + a₀, the leading coefficient is a_n.

Your polynomial f(x) has zeros at -2, 3, 6. So f(-2) = f(3) = f(6) = 0.

This means that (x + 2), (x - 3), and (x - 6) are factors of f(x).

If a binomial "x + 2" is a factor of f(x), f(x) can be written as a product of (x + 2) and something else. Because f(x) is a product of (x + 2) and something, if x + 2 = 0 (because x = -2), then f(x) = 0 times something = 0.

The polynomial of the smallest degree with (x+2), (x-3), and (x-6) as factors would look like

f(x) = a(x+2)(x-3)(x-6). Well your leading coefficient was 1, so this "a" must be 1. Otherwise your x³ term will have a coefficient that's not 1.

Now all we need to do is expand f(x) = (x+2)(x-3)(x-6) so that it becomes a standard form.

(x+2)(x-3) = x²-x-6

(x²-x-6)(x-6) = x³ - 6x² - x² + 6x - 6x + 36 = x³ - 7x² + 36

f(x) = x³ - 7x² + 36.