Solve a^3-4*(0)^2+2*(0)-1=

$$a ^ { 3 } - 4 \cdot ( 0 ) ^ { 2 } + 2 \cdot ( 0 ) - 1 =$$

Evaluate

$$a^{3}-1$$

Solution Steps

Calculate 0 to the power of 2 and get 0.

$$a^{3}-4\times 0+2\times 0-1$$

Multiply 4 and 0 to get 0.

$$a^{3}-0+2\times 0-1$$

Multiply 2 and 0 to get 0.

$$a^{3}-0+0-1$$

Subtract 1 from 0 to get -1.

$$a^{3}-0-1$$

Multiply -1 and 0 to get 0.

$$a^{3}+0-1$$

Anything plus zero gives itself.

$$a^{3}-1$$

Factor

$$\left(a-1\right)\left(a^{2}+a+1\right)$$

Solution Steps

Multiply and combine like terms.

$$a^{3}-1$$

Rewrite a^{3}-1 as a^{3}-1^{3}. The difference of cubes can be factored using the rule: p^{3}-q^{3}=\left(p-q\right)\left(p^{2}+pq+q^{2}\right). Polynomial a^{2}+a+1 is not factored since it does not have any rational roots.

$$\left(a-1\right)\left(a^{2}+a+1\right)$$