Question

solve the equation 16x^3 + 16x^2 - x - 1 = 0 given that -\frac{1}{4} is a zero of f(x)=16x^3 + 16x^2 - x - 1. the solution set is { }. (use a comma to separate answers as needed.)

Explanation:

Step1: Use factor - theorem

Since $-\frac{1}{4}$ is a zero of $f(x)=16x^{3}+16x^{2}-x - 1$, then $(x+\frac{1}{4})$ is a factor of $16x^{3}+16x^{2}-x - 1$. We use long - division or synthetic division. Let's use synthetic division. The coefficients of the polynomial are $16,16, - 1,-1$.
The synthetic - division setup:

-1/4 |  16  16  -1  -1
      |     -4  -3   1
      |----------------
       16  12  -4   0

The quotient is $16x^{2}+12x - 4$.

Step2: Factor the quotient

Factor out the greatest common factor from $16x^{2}+12x - 4$. We can factor out $4$ first: $4(4x^{2}+3x - 1)$.
Then factor $4x^{2}+3x - 1$ using the AC - method. We need to find two numbers that multiply to $4\times(-1)=-4$ and add up to $3$. The numbers are $4$ and $-1$.
$4x^{2}+3x - 1=4x^{2}+4x - x - 1=4x(x + 1)-(x + 1)=(4x - 1)(x + 1)$
So, $16x^{3}+16x^{2}-x - 1=(x+\frac{1}{4})(4x - 1)(x + 1)$

Step3: Solve for x

Set each factor equal to zero:
$x+\frac{1}{4}=0\Rightarrow x=-\frac{1}{4}$
$4x - 1=0\Rightarrow4x=1\Rightarrow x=\frac{1}{4}$
$x + 1=0\Rightarrow x=-1$

Answer:

$-\frac{1}{4},\frac{1}{4},-1$