# what are the other zeros of f(x)=3x^3+17x^2+18x-8 if one zero is x=-4

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You may write the polynomial in factored form such that:

`f(x) = (x-(-4))(ax^2 + bx + c)`

`3x^3+17x^2+18x-8 =(x+4)(ax^2 + bx + c)`

Notice the quadratic form of quotient, `ax^2 + bx + c` , and the absence of reminder (the reminder theorem tells that if x=a is a zero for polynomial f(x), then the reminder is zero).

Opening the brackets to the right yields:

`3x^3+17x^2+18x-8 =ax^3 + bx^2 + cx + 4ax^2 + 4bx + 4c`

Grouping the terms containing like powers of x yields:

`3x^3+17x^2+18x-8 =ax^3 +x^2*(b+4a) + x*(c + 4b) + 4c`

Equating the like powers of x both sides yields:

a = 3

b + 4a = 17 => b + 12 = 17 => b = 5

c + 4b = 18 => c = 18 - 20 => c = -2

This checks the last equation 4c=-8 => c = -2

Hence, the quotient is `q(x) = 3x^2 + 5x - 2`

You need to find the next two zeroes of polynomial f(x), hence `(x+4)(3x^2 + 5x - 2) = 0 =gt 3x^2 + 5x - 2 = 0` .

You may use quadratic formula to find the two zeroes of q(x).

`x_(1,2) = (-5+-sqrt(25 + 24))/6=gtx_(1,2) = (-5+-sqrt49)/6` `x_(1,2) = (-5+-7)/6=gtx_1 = 1/3; x_2 = -2`

**Hence, the rest of the zeroes of f(x) determined using reminder theorem are `x_1=1/3 ; x_2 = -2` .**

Given `f(x)=3x^3+17x^2+18x-8` and `f(-4)=0` , find the remaining zeros.

(1) If `x=-4` is a zero, then `(x+4)` is a factor. Then we can use long division :

` ` `3x^2+5x-2`

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

`x+4` | `3x^3+17x^2+18x-8`

`3x^3+12x^2`

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

`5x^2+18x`

`5x^2+20x`

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

`-2x-8`

`-2x-8`

--------

0

(2) Thus `3x^3+17x^2+18x-8=(x+4)(3x^2+5x-2)`

The quadratic factors yielding:

`3x^3+17x^2+18x-8=(x+4)(x+2)(3x-1)`

Thus if `(x+4)(x+2)(3x-1)=0` by the zero product property we have:

`x=-4,x=-2,"or" x=1/3`

**So the remaining zeros are -2 and 1/3.**