One zero of f(x)=x^3+4x^2+x+c is -3.
a) calculate the value of c.
b)calculate the other zeros.
Ummm, didn't MathMate just do one of these for you? Plug in -3 and evaluate to find c.
Ummmmm, I needed help with the second part
Divide by x+3 and see what you get:
x^3 + 4x^2 + x -6
= (x+3)(x^2 + x - 2)
= (x+3)(x+2)(x-1)
Got it now?
what about this question:
Determine the roots algebraically by factoring.
a) x^3-8x^2-3x+90=0
To find the value of c, we can substitute the given zero into the equation and solve for c.
a) To do this, we insert the given zero (-3) into the equation:
f(-3) = (-3)^3 + 4(-3)^2 + (-3) + c
Simplifying:
f(-3) = -27 + 36 - 3 + c
Combining like terms:
f(-3) = 6 + c
Since we know that -3 is a zero of f(x), this means that f(-3) = 0. So we can set it equal to zero:
0 = 6 + c
Now, we can solve for c:
c = -6
Therefore, the value of c is -6.
b) To find the other zeros, we can use the fact that the sum of the zeros of a cubic function is equal to the negative coefficient of the quadratic term (in this case, 4).
Since we know one zero is -3, we can use this information to find the other zeros.
The sum of the zeros is equal to the sum of all three zeros, which is (-3) + x + y, where x and y are the other two zeros.
According to the fact mentioned above, the sum of the zeros is also equal to -4 (the negative coefficient of the quadratic term).
Therefore, (-3) + x + y = -4.
To find the other two zeros, we need to solve this equation. Since we only have one equation, we need additional information to fully determine the other zeros.