can someone help me find the zeros of these functions algebraically? thanks its greatly appreciated?

Question

find the zeros of these functions algebraically

f(x) = 2x^2-7x-30
f(x) = x^2-9x+14
f(x) = x^3 - 4x^2-9x+36

Answer 1

2x^2-7x-30=0
(2x+5)(x-6)=0
2x+5=0 or x-6=0
x=5/2 or x=6
------------------------------
x^2-9x+14=0
(x-7)(x-2)=0
x=7 or x=2
---------------
x^3 - 4x^2-9x+36=0
x^2(x-4)-9(x-4)=0
(x-4)(x^2-9)=0
(x-4)(x-3)(x+3)=0
x=4 ,x=3 ,x=-3

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Answer 2

The first two are quadratic equations of the form ax²+bx+c=0. The solutions for these are
(-b +- sqrt(b²-4ac))/2a
1) (7 +- sqrt(289))/4 = (7 +- 17)/4 = -2,5 and 6
2) (9 +- sqrt(25)/2 = (9 +- 5)/2 = 2 and 7
The third is a cubic equation and the solution to those is far more complex. I can tell you that the solutions are -3, 3 and 4 but the formula is too complex to give.

Answer 3

you should try to rewrite them as product of two linear polynomes. but if you cant you can always use the abc-formula.

f(x) = x^2-9x+14

find two numbers a and b with
a*b = 14
a+b=-9

a = -7 b = -2 does the trick

sooooo : x^2-9x+14 = (x+7)(x+2)
the zeros are x = -7, and x = -2

Answer 4

(i) 2x^2 - 7x - 30 = 0, we can factorise to obtain
(2x +5)(x - 6)=0, so
Either
2x+5=0
2x=-5
x=5/2
x=2.5

or
x-6=0
x=6

(ii) x^2-9x+14=0, factorising, gives
(x-7)(x-2)=0
So either
x-7=0
x=7

or
x-2=0
x=2


(iii)x^3-4x^2-9x+36=0
(x+3)(x-3)(x-4)=0 on factorising
Therefore, either
x+3=0
x=-3

or
x-3=0
x=3

or
x-4=0
x=4
Hope this helps.

Answer 5

Do your own homework so that you can learn something.