Find the Roots (Zeros) x^5-3x^4-15x^3+45x^2-16x+48=0

$x^{5} - 3 x^{4} - 15 x^{3} + 45 x^{2} - 16 x + 48 = 0$

Step 1

Factor the left side of the equation.

Regroup terms.

$x^{5} - 15 x^{3} - 16 x - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor $x$ out of $x^{5} - 15 x^{3} - 16 x$ .

Factor $x$ out of $x^{5}$ .

$x \cdot x^{4} - 15 x^{3} - 16 x - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor $x$ out of $- 15 x^{3}$ .

$x \cdot x^{4} + x (- 15 x^{2}) - 16 x - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor $x$ out of $- 16 x$ .

$x \cdot x^{4} + x (- 15 x^{2}) + x \cdot - 16 - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor $x$ out of $x \cdot x^{4} + x (- 15 x^{2})$ .

$x (x^{4} - 15 x^{2}) + x \cdot - 16 - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor $x$ out of $x (x^{4} - 15 x^{2}) + x \cdot - 16$ .

$x (x^{4} - 15 x^{2} - 16) - 3 x^{4} + 45 x^{2} + 48 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$x ({(x^{2})}^{2} - 15 x^{2} - 16) - 3 x^{4} + 45 x^{2} + 48 = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$x (u^{2} - 15 u - 16) - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor $u^{2} - 15 u - 16$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 16$ and whose sum is $- 15$ .

$- 16, 1$

Write the factored form using these integers.

$x ((u - 16) (u + 1)) - 3 x^{4} + 45 x^{2} + 48 = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$x ((x^{2} - 16) (x^{2} + 1)) - 3 x^{4} + 45 x^{2} + 48 = 0$

Rewrite $16$ as $4^{2}$ .

$x ((x^{2} - 4^{2}) (x^{2} + 1)) - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

$x ((x + 4) (x - 4) (x^{2} + 1)) - 3 x^{4} + 45 x^{2} + 48 = 0$

Remove unnecessary parentheses.

$x (x + 4) (x - 4) (x^{2} + 1) - 3 x^{4} + 45 x^{2} + 48 = 0$

Factor $3$ out of $- 3 x^{4} + 45 x^{2} + 48$ .

Factor $3$ out of $- 3 x^{4}$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- x^{4}) + 45 x^{2} + 48 = 0$

Factor $3$ out of $45 x^{2}$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- x^{4}) + 3 (15 x^{2}) + 48 = 0$

Factor $3$ out of $48$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- x^{4}) + 3 (15 x^{2}) + 3 (16) = 0$

Factor $3$ out of $3 (- x^{4}) + 3 (15 x^{2})$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- x^{4} + 15 x^{2}) + 3 (16) = 0$

Factor $3$ out of $3 (- x^{4} + 15 x^{2}) + 3 (16)$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- x^{4} + 15 x^{2} + 16) = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- {(x^{2})}^{2} + 15 x^{2} + 16) = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- u^{2} + 15 u + 16) = 0$

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 $a \cdot c = - 1 \cdot 16 = - 16$ and whose sum is $b = 15$ .

Factor $15$ out of $15 u$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- u^{2} + 15 (u) + 16) = 0$

Rewrite $15$ as $- 1$ plus $16$

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- u^{2} + (- 1 + 16) u + 16) = 0$

Apply the distributive property.

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (- u^{2} - 1 u + 16 u + 16) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x (x + 4) (x - 4) (x^{2} + 1) + 3 ((- u^{2} - 1 u) + 16 u + 16) = 0$

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

$x (x + 4) (x - 4) (x^{2} + 1) + 3 (u (- u - 1) - 16 (- u - 1)) = 0$

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

$x (x + 4) (x - 4) (x^{2} + 1) + 3 ((- u - 1) (u - 16)) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$x (x + 4) (x - 4) (x^{2} + 1) + 3 ((- x^{2} - 1) (x^{2} - 16)) = 0$

Rewrite $16$ as $4^{2}$ .

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

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 4$ .

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

Remove unnecessary parentheses.

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

Remove unnecessary parentheses.

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

Factor $(x + 4) (x - 4)$ out of $x (x + 4) (x - 4) (x^{2} + 1) + 3 (- x^{2} - 1) (x + 4) (x - 4)$ .

Factor $(x + 4) (x - 4)$ out of $x (x + 4) (x - 4) (x^{2} + 1)$ .

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

Factor $(x + 4) (x - 4)$ out of $3 (- x^{2} - 1) (x + 4) (x - 4)$ .

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

Factor $(x + 4) (x - 4)$ out of $(x + 4) (x - 4) ((x) (x^{2} + 1)) + (x + 4) (x - 4) (3 (- x^{2} - 1))$ .

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

Apply the distributive property.

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

Multiply $x$ by $x^{2}$ by adding the exponents.

Multiply $x$ by $x^{2}$ .

Raise $x$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $2$ .

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

Multiply $x$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $3$ .

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

Multiply $3$ by $- 1$ .

$(x + 4) (x - 4) (x^{3} + x - 3 x^{2} - 3) = 0$

Reorder terms.

$(x + 4) (x - 4) (x^{3} - 3 x^{2} + x - 3) = 0$

Factor.

Rewrite $x^{3} - 3 x^{2} + x - 3$ in a factored form.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x + 4) (x - 4) ((x^{3} - 3 x^{2}) + x - 3) = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $x - 3$ .

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

Remove unnecessary parentheses.

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

Step 2

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

$x + 4 = 0$

$x - 4 = 0$

$x - 3 = 0$

$x^{2} + 1 = 0$

Step 3

Set $x + 4$ equal to $0$ and solve for $x$ .

Set $x + 4$ equal to $0$ .

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 4

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 5

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 6

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{- 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = i$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - i$

The complete solution is the result of both the positive and negative portions of the solution.

$x = i, - i$

Step 7

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

$x = - 4, 4, 3, i, - i$

Step 8

Find the Roots (Zeros) x^5-3x^4-15x^3+45x^2-16x+48=0

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