Solve by Factoring 4k^(4/3)-17k^(2/3)+4=0

$4 k^{\frac{4}{3}} - 17 k^{\frac{2}{3}} + 4 = 0$

Rewrite $k^{\frac{4}{3}}$ as ${(k^{\frac{2}{3}})}^{2}$ .

$4 {(k^{\frac{2}{3}})}^{2} - 17 k^{\frac{2}{3}} + 4 = 0$

Let $u = k^{\frac{2}{3}}$ . Substitute $u$ for all occurrences of $k^{\frac{2}{3}}$ .

$4 u^{2} - 17 u + 4 = 0$

Factor by grouping.

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For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 4 = 16$ and whose sum is $b = - 17$ .

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Factor $- 17$ out of $- 17 u$ .

$4 u^{2} - 17 u + 4 = 0$

Rewrite $- 17$ as $- 1$ plus $- 16$

$4 u^{2} + (- 1 - 16) u + 4 = 0$

Apply the distributive property.

$4 u^{2} - 1 u - 16 u + 4 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) - 16 u + 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (4 u - 1) - 4 (4 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$(4 u - 1) (u - 4) = 0$

Replace all occurrences of $u$ with $k^{\frac{2}{3}}$ .

$(4 k^{\frac{2}{3}} - 1) (k^{\frac{2}{3}} - 4) = 0$

Rewrite $4 k^{\frac{2}{3}}$ as ${(2 k^{\frac{1}{3}})}^{2}$ .

$({(2 k^{\frac{1}{3}})}^{2} - 1) (k^{\frac{2}{3}} - 4) = 0$

Rewrite $1$ as $1^{2}$ .

$({(2 k^{\frac{1}{3}})}^{2} - 1^{2}) (k^{\frac{2}{3}} - 4) = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2 k^{\frac{1}{3}}$ and $b = 1$ .

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) (k^{\frac{2}{3}} - 4) = 0$

Rewrite $k^{\frac{2}{3}}$ as ${(k^{\frac{1}{3}})}^{2}$ .

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) ({(k^{\frac{1}{3}})}^{2} - 4) = 0$

Rewrite $4$ as $2^{2}$ .

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) ({(k^{\frac{1}{3}})}^{2} - 2^{2}) = 0$

Factor.

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Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = k^{\frac{1}{3}}$ and $b = 2$ .

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) ((k^{\frac{1}{3}} + 2) (k^{\frac{1}{3}} - 2)) = 0$

Remove unnecessary parentheses.

$(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) (k^{\frac{1}{3}} + 2) (k^{\frac{1}{3}} - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 k^{\frac{1}{3}} + 1 = 0$

$2 k^{\frac{1}{3}} - 1 = 0$

$k^{\frac{1}{3}} + 2 = 0$

$k^{\frac{1}{3}} - 2 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$2 k^{\frac{1}{3}} + 1 = 0$

Subtract $1$ from both sides of the equation.

$2 k^{\frac{1}{3}} = - 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 k^{\frac{1}{3}} = - 1$ by $2$ .

$\frac{2 k^{\frac{1}{3}}}{2} = \frac{- 1}{2}$

Simplify $\frac{2 k^{\frac{1}{3}}}{2}$ .

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Cancel the common factor.

$\frac{2 k^{\frac{1}{3}}}{2} = \frac{- 1}{2}$

Divide $k^{\frac{1}{3}}$ by $1$ .

$k^{\frac{1}{3}} = \frac{- 1}{2}$

Move the negative in front of the fraction.

$k^{\frac{1}{3}} = - \frac{1}{2}$

Raise each side of the equation to the $3$ power to eliminate the fractional exponent on the left side.

${(k^{\frac{1}{3}})}^{3} = {(- \frac{1}{2})}^{3}$

Simplify ${(- \frac{1}{2})}^{3}$ .

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Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $- \frac{1}{2}$ .

$k = {(- 1)}^{3} {(\frac{1}{2})}^{3}$

Apply the product rule to $\frac{1}{2}$ .

$k = {(- 1)}^{3} \frac{1^{3}}{2^{3}}$

Raise $- 1$ to the power of $3$ .

$k = - \frac{1^{3}}{2^{3}}$

One to any power is one.

$k = - \frac{1}{2^{3}}$

Raise $2$ to the power of $3$ .

$k = - \frac{1}{8}$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$2 k^{\frac{1}{3}} - 1 = 0$

Add $1$ to both sides of the equation.

$2 k^{\frac{1}{3}} = 1$

Divide each term by $2$ and simplify.

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Divide each term in $2 k^{\frac{1}{3}} = 1$ by $2$ .

$\frac{2 k^{\frac{1}{3}}}{2} = \frac{1}{2}$

Simplify $\frac{2 k^{\frac{1}{3}}}{2}$ .

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Cancel the common factor.

$\frac{2 k^{\frac{1}{3}}}{2} = \frac{1}{2}$

Divide $k^{\frac{1}{3}}$ by $1$ .

$k^{\frac{1}{3}} = \frac{1}{2}$

Raise each side of the equation to the $3$ power to eliminate the fractional exponent on the left side.

${(k^{\frac{1}{3}})}^{3} = {(\frac{1}{2})}^{3}$

Simplify ${(\frac{1}{2})}^{3}$ .

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Apply the product rule to $\frac{1}{2}$ .

$k = \frac{1^{3}}{2^{3}}$

One to any power is one.

$k = \frac{1}{2^{3}}$

Raise $2$ to the power of $3$ .

$k = \frac{1}{8}$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$k^{\frac{1}{3}} + 2 = 0$

Subtract $2$ from both sides of the equation.

$k^{\frac{1}{3}} = - 2$

Raise each side of the equation to the $3$ power to eliminate the fractional exponent on the left side.

${(k^{\frac{1}{3}})}^{3} = {(- 2)}^{3}$

Raise $- 2$ to the power of $3$ .

$k = - 8$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$k^{\frac{1}{3}} - 2 = 0$

Add $2$ to both sides of the equation.

$k^{\frac{1}{3}} = 2$

Raise each side of the equation to the $3$ power to eliminate the fractional exponent on the left side.

${(k^{\frac{1}{3}})}^{3} = {(2)}^{3}$

Raise $2$ to the power of $3$ .

$k = 8$

The final solution is all the values that make $(2 k^{\frac{1}{3}} + 1) (2 k^{\frac{1}{3}} - 1) (k^{\frac{1}{3}} + 2) (k^{\frac{1}{3}} - 2) = 0$ true.

$k = - \frac{1}{8}, \frac{1}{8}, - 8, 8$

Solve by Factoring 4k^(4/3)-17k^(2/3)+4=0

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