Find all real zeros for each equation.
1. f(x)=x^3-3x^2-4x
2. f(x)= x^6-x^2
3. f(x)=x^3+3x^2-16x-48
4. f(x)=9x^4-37x^2+4
I will do one, then you try
x (x^2 - 3 x - 4) = 0
x (x-4)(x+1) = 0
so x = 0 , 4 , -1
1. take out common factor of x, the rest factors
2. factor out x^2, then difference of squares.
one of them goes again.
3. x^3 + 3x^2 - 16x - 48
looks like "grouping" works here
= x^2(x+3) - 16(x+3)
= (x+3)(x^2- 16) , and a difference of squares
=(x+3)(x+4)(x-4)
4. sneaky one ...
9x^4-37x^2+4
= (x^2 - 1)(x^2 - 4)
= ..... can you finish it ?
last one should be
4. sneaky one ...
9x^4-37x^2+4
= (9x^2 - 1)(x^2 - 4)
= ..... can you finish it ?
To find the real zeros of a polynomial equation, we set the equation equal to zero and solve for x. Let's go through each equation and find its real zeros.
1. f(x) = x^3 - 3x^2 - 4x
To find the real zeros of this equation, we can use the Rational Root Theorem. According to the theorem, any rational root of the equation must be a factor of the constant term (in this case, -4) divided by a factor of the leading coefficient (in this case, 1).
The possible rational roots are ±1, ±2, and ±4.
For each possible rational root, we substitute it into the equation and see if the result is zero. By trying these values, we find that x = -1 and x = 4 are the real zeros of the equation.
Therefore, the real zeros for this equation are x = -1 and x = 4.
2. f(x) = x^6 - x^2
This equation can be factored as f(x) = x^2(x^4 - 1). We can further factor the second term as a difference of squares to get f(x) = x^2(x^2 - 1)(x^2 + 1).
Setting each factor equal to zero, we find x = 0, x = 1, and x = -1 as the real zeros of the equation.
Therefore, the real zeros for this equation are x = 0, x = 1, and x = -1.
3. f(x) = x^3 + 3x^2 - 16x - 48
Again, we can use the Rational Root Theorem to find the possible rational roots of this equation. The possible rational roots are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48.
By substituting these values, we find that x = -4 and x = 3 are the real zeros of the equation.
Therefore, the real zeros for this equation are x = -4 and x = 3.
4. f(x) = 9x^4 - 37x^2 + 4
We notice that this equation is a quadratic equation in disguise. We can rewrite it as f(x) = (3x^2 - 1)(3x^2 - 4).
Setting each factor equal to zero, we find x = ±1/√3 and x = ±2/√3 as the real zeros of the equation.
Therefore, the real zeros for this equation are x = ±1/√3 and x = ±2/√3.