Solve k[k(-1)^2-2(k+4)(-1)-k+16]0

$$k [ k ( - 1 ) ^ { 2 } - 2 ( k + 4 ) ( - 1 ) - k + 16 ] 0$$

Evaluate

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Solution Steps

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

$$k\left(k\times 1-2\left(k+4\right)\left(-1\right)-k+16\right)\times 0$$

Multiply 2 and -1 to get -2.

$$k\left(k\times 1-\left(-2\left(k+4\right)\right)-k+16\right)\times 0$$

The opposite of -2\left(k+4\right) is 2\left(k+4\right).

$$k\left(k\times 1+2\left(k+4\right)-k+16\right)\times 0$$

Use the distributive property to multiply 2 by k+4.

$$k\left(k\times 1+2k+8-k+16\right)\times 0$$

Combine k\times 1 and 2k to get 3k.

$$k\left(3k+8-k+16\right)\times 0$$

Combine 3k and -k to get 2k.

$$k\left(2k+8+16\right)\times 0$$

Add 8 and 16 to get 24.

$$k\left(2k+24\right)\times 0$$

Anything times zero gives zero.

$$0$$

Factor

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