Find the zeros of the function by factoring or algebraically.?

Question

f(x)=7x^4+14x^3-168x^2

Answer 1

set the equation equals to 0

0 = 7x^4+14x^3-168x^2

factor out 7x^2
0 = 7x^2 (x^2 + 2x - 24)

factor again. find two numbers that multiply to -24 and add to 2.
The two numbers are 6 and -4
0 = 7x^2 (x + 6) (x - 4)

x = 0 -6 or 4

Answer 2

Notice that each term is divisible by 7x²:

7x^4+14x³-168x² = 7x²(x²+2x-24)

Then factor the term in parentheses:

7x²(x²+2x-24) = 7x²(x+6)(x-4)

To determine zeroes, set f(x)=0:

7x²(x+6)(x-4)=0
7x²=0 or x+6=0 or x-4=0
x=0 or x=-6 or x=4

So your zeroes are 0, -6, and 4.

Answer 3

f(x) = 7x^4+14x^3-168x^2

factor 7x^2 ---> f(x) = 7x^2*(x^2+2x - 24)

Set f(x) = 0 = 7x^2*(x^2+2x-24) ok one zero is x = 0

So now factor (x^2+2x-24)

(x^2+2x-24) = (x+6)(x-4) ---> zeros at x = -6 and x=4

so zeros are: x= -6, 0, 0, 4 Zero is mentioned twice since there are 4 roots to a fourth order equation and the zero comes from x^2 = 0, hence a degenerate (repeated) root.

Answer 4

7x^4 + 14x^3 - 168x^2 = 0
7x²(x² + 2x - 24) = 0
7x²(x + 6)(x - 4) = 0

x = -6, 0, 4