Solve (k-3)(k^(2) 3k 9)

Question

Simplify the expression

27 k^{4} - 81 k^{3}

Evaluate

(k - 3) (k^{2} \times 3 k \times 9)

Remove the parentheses

(k - 3) k^{2} \times 3 k \times 9

Multiply the terms with the same base by adding their exponents

(k - 3) k^{2 + 1} \times 3 \times 9

Add the numbers

(k - 3) k^{3} \times 3 \times 9

Multiply the terms

(k - 3) k^{3} \times 27

Use the commutative property to reorder the terms

(k - 3) \times 27 k^{3}

Multiply the terms

27 k^{3} (k - 3)

Apply the distributive property

27 k^{3} \times k - 27 k^{3} \times 3

Multiply the terms

More Steps

Evaluate

k^{3} \times k

Use the product rule a^{n} \times a^{m} = a^{n + m} to simplify the expression

k^{3 + 1}

Add the numbers

k^{4}

27 k^{4} - 27 k^{3} \times 3

Solution

27 k^{4} - 81 k^{3}

Show Solution

k_{1} = 0, k_{2} = 3

Evaluate

(k - 3) (k^{2} \times 3 k \times 9)

To find the roots of the expression,set the expression equal to 0

(k - 3) (k^{2} \times 3 k \times 9) = 0

Multiply

More Steps

Multiply the terms

k^{2} \times 3 k \times 9

Multiply the terms with the same base by adding their exponents

k^{2 + 1} \times 3 \times 9

Add the numbers

k^{3} \times 3 \times 9

Multiply the terms

k^{3} \times 27

Use the commutative property to reorder the terms

27 k^{3}

(k - 3) \times 27 k^{3} = 0

Multiply the terms

27 k^{3} (k - 3) = 0

Elimination the left coefficient

k^{3} (k - 3) = 0

Separate the equation into 2 possible cases

k^{3} = 0 k - 3 = 0

The only way a power can be 0 is when the base equals 0

k = 0 k - 3 = 0

Solve the equation

More Steps

Evaluate

k - 3 = 0

Move the constant to the right-hand side and change its sign

k = 0 + 3

Removing 0 doesn't change the value,so remove it from the expression

k = 3

k = 0 k = 3

Solution

k_{1} = 0, k_{2} = 3

Show Solution