Hurry 98 which polynomial function has a leading coefficient of 1, roots –2 and 7 with multiplicity 1, and root 5 with multiplicity 2? f(x) = 2(x + 7)(x + 5)(x – 2) f(x) = 2(x – 7)(x – 5)(x + 2) f(x) = (x + 7)(x + 5)(x + 5)(x – 2) f(x) = (x – 7)(x – 5)(x – 5)(x + 2)

f(x) = (x – 7)(x – 5)(x – 5)(x + 2) should be your answer

To get the x, -2, 7, 5 and 5, you must first place each parenthesis = to zero. Solve for x.

x - 7 = 0, x - 7 (+7) = 0 (+7), x = 7

x - 5 = 0, x - 5 (+5) = 0 (+5), x = 5

x - 5 = 0, x - 5 (+5) = 0 (+5), x = 5

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

hope this helps

Hello there!

We are looking for a polinomyal with a leading coefficient of 1.

This means that the coefficient in the highest order power is equal to 1, with this information we can discard options 1 and 2, because both of them  have a leading coefficient of 2.

now, we know that the roots are -2, 7, and two times 5 (because has multiplicity of 2)

so we have 4 roots, this means that the polynomail is of order 4, and we can write it as:

f(x) = (x - root 1)(x - root 2)(x - root 3)(x- root 4)

in this case, we have:

f(x) = (x +2)(x - 7)(x-5)(x-5), we can reorder this and get:

f(x) = (x-7)(x-5)(x-5)(x+2)

wich is the fourth option, then the right answer is the fourth option.

step-by-step explanation:

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