Solve p(x)=4(x)^(3)-6(x)^(2)+3x+14,x=-1 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$p(x)=4 { x }^{ 3 } -6 { x }^{ 2 } +3x+14,x=-1$$

Solve for p, x

$x=-1$
$p=-1$

Solution Steps

Consider the first equation. Insert the known values of variables into the equation.

$$p\left(-1\right)=4\left(-1\right)^{3}-6\left(-1\right)^{2}+3\left(-1\right)+14$$

Calculate $-1$ to the power of $3$ and get $-1$.

$$p\left(-1\right)=4\left(-1\right)-6\left(-1\right)^{2}+3\left(-1\right)+14$$

Multiply $4$ and $-1$ to get $-4$.

$$p\left(-1\right)=-4-6\left(-1\right)^{2}+3\left(-1\right)+14$$

Calculate $-1$ to the power of $2$ and get $1$.

$$p\left(-1\right)=-4-6+3\left(-1\right)+14$$

Subtract $6$ from $-4$ to get $-10$.

$$p\left(-1\right)=-10+3\left(-1\right)+14$$

Multiply $3$ and $-1$ to get $-3$.

$$p\left(-1\right)=-10-3+14$$

Subtract $3$ from $-10$ to get $-13$.

$$p\left(-1\right)=-13+14$$

Add $-13$ and $14$ to get $1$.

$$p\left(-1\right)=1$$

Divide both sides by $-1$.

$$p=-1$$

The system is now solved.

$$p=-1$$ $$x=-1$$