Solve by Factoring 6x^(2/3)-7x^(1/3)-20=0

$6 x^{\frac{2}{3}} - 7 x^{\frac{1}{3}} - 20 = 0$

Step 1

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

$6 {(x^{\frac{1}{3}})}^{2} - 7 x^{\frac{1}{3}} - 20 = 0$

Step 2

Let $u = x^{\frac{1}{3}}$ . Substitute $u$ for all occurrences of $x^{\frac{1}{3}}$ .

$6 u^{2} - 7 u - 20 = 0$

Step 3

Factor by grouping.

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 $a \cdot c = 6 \cdot - 20 = - 120$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 u$ .

$6 u^{2} - 7 u - 20 = 0$

Rewrite $- 7$ as $8$ plus $- 15$

$6 u^{2} + (8 - 15) u - 20 = 0$

Apply the distributive property.

$6 u^{2} + 8 u - 15 u - 20 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(6 u^{2} + 8 u) - 15 u - 20 = 0$

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

$2 u (3 u + 4) - 5 (3 u + 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 u + 4$ .

$(3 u + 4) (2 u - 5) = 0$

Step 4

Replace all occurrences of $u$ with $x^{\frac{1}{3}}$ .

$(3 x^{\frac{1}{3}} + 4) (2 x^{\frac{1}{3}} - 5) = 0$

Step 5

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

$3 x^{\frac{1}{3}} + 4 = 0$

$2 x^{\frac{1}{3}} - 5 = 0$

Step 6

Set $3 x^{\frac{1}{3}} + 4$ equal to $0$ and solve for $x$ .

Set $3 x^{\frac{1}{3}} + 4$ equal to $0$ .

$3 x^{\frac{1}{3}} + 4 = 0$

Solve $3 x^{\frac{1}{3}} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$3 x^{\frac{1}{3}} = - 4$

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

${(3 x^{\frac{1}{3}})}^{3} = {(- 4)}^{3}$

Simplify the exponent.

Simplify the left side.

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

Apply the product rule to $3 x^{\frac{1}{3}}$ .

$3^{3} {(x^{\frac{1}{3}})}^{3} = {(- 4)}^{3}$

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

$27 {(x^{\frac{1}{3}})}^{3} = {(- 4)}^{3}$

Multiply the exponents in ${(x^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$27 x^{\frac{1}{3} \cdot 3} = {(- 4)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$27 x^{\frac{1}{3} \cdot 3} = {(- 4)}^{3}$

Rewrite the expression.

$27 x^{1} = {(- 4)}^{3}$

Simplify.

$27 x = {(- 4)}^{3}$

Simplify the right side.

Raise $- 4$ to the power of $3$ .

$27 x = - 64$

Divide each term in $27 x = - 64$ by $27$ and simplify.

Divide each term in $27 x = - 64$ by $27$ .

$\frac{27 x}{27} = \frac{- 64}{27}$

Simplify the left side.

Cancel the common factor of $27$ .

Cancel the common factor.

$\frac{27 x}{27} = \frac{- 64}{27}$

Divide $x$ by $1$ .

$x = \frac{- 64}{27}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{64}{27}$

Step 7

Set $2 x^{\frac{1}{3}} - 5$ equal to $0$ and solve for $x$ .

Set $2 x^{\frac{1}{3}} - 5$ equal to $0$ .

$2 x^{\frac{1}{3}} - 5 = 0$

Solve $2 x^{\frac{1}{3}} - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$2 x^{\frac{1}{3}} = 5$

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

${(2 x^{\frac{1}{3}})}^{3} = 5^{3}$

Simplify the exponent.

Simplify the left side.

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

Apply the product rule to $2 x^{\frac{1}{3}}$ .

$2^{3} {(x^{\frac{1}{3}})}^{3} = 5^{3}$

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

$8 {(x^{\frac{1}{3}})}^{3} = 5^{3}$

Multiply the exponents in ${(x^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$8 x^{\frac{1}{3} \cdot 3} = 5^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$8 x^{\frac{1}{3} \cdot 3} = 5^{3}$

Rewrite the expression.

$8 x^{1} = 5^{3}$

Simplify.

$8 x = 5^{3}$

Simplify the right side.

Raise $5$ to the power of $3$ .

$8 x = 125$

Divide each term in $8 x = 125$ by $8$ and simplify.

Divide each term in $8 x = 125$ by $8$ .

$\frac{8 x}{8} = \frac{125}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{125}{8}$

Divide $x$ by $1$ .

$x = \frac{125}{8}$

Step 8

The final solution is all the values that make $(3 x^{\frac{1}{3}} + 4) (2 x^{\frac{1}{3}} - 5) = 0$ true.

$x = - \frac{64}{27}, \frac{125}{8}$

Solve by Factoring 6x^(2/3)-7x^(1/3)-20=0

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