Find the x and y Intercepts y=2x^4+8x^3+4x^2-8x-6

$y = 2 x^{4} + 8 x^{3} + 4 x^{2} - 8 x - 6$

Step 1

Find the x-intercepts.

To find the x-intercept(s), substitute in $0$ for $y$ and solve for $x$ .

$0 = 2 x^{4} + 8 x^{3} + 4 x^{2} - 8 x - 6$

Solve the equation.

Rewrite the equation as $2 x^{4} + 8 x^{3} + 4 x^{2} - 8 x - 6 = 0$ .

$2 x^{4} + 8 x^{3} + 4 x^{2} - 8 x - 6 = 0$

Factor the left side of the equation.

Factor $2$ out of $2 x^{4} + 8 x^{3} + 4 x^{2} - 8 x - 6$ .

Factor $2$ out of $2 x^{4}$ .

$2 (x^{4}) + 8 x^{3} + 4 x^{2} - 8 x - 6 = 0$

Factor $2$ out of $8 x^{3}$ .

$2 (x^{4}) + 2 (4 x^{3}) + 4 x^{2} - 8 x - 6 = 0$

Factor $2$ out of $4 x^{2}$ .

$2 (x^{4}) + 2 (4 x^{3}) + 2 (2 x^{2}) - 8 x - 6 = 0$

Factor $2$ out of $- 8 x$ .

$2 (x^{4}) + 2 (4 x^{3}) + 2 (2 x^{2}) + 2 (- 4 x) - 6 = 0$

Factor $2$ out of $- 6$ .

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

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

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

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

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

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

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

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

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

Regroup terms.

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

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

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

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

Factor $4 x$ out of $- 4 x$ .

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

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

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

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

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

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 $b = 1$ .

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

Remove unnecessary parentheses.

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

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

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

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

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

Factor $u^{2} + 2 u - 3$ 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 $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$2 (4 x (x + 1) (x - 1) + (u - 1) (u + 3)) = 0$

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

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

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

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

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 $b = 1$ .

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

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

Factor $(x + 1) (x - 1)$ out of $4 x (x + 1) (x - 1)$ .

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

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

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

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

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

Factor $4 u + u^{2} + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$2 ((x + 1) (x - 1) ((u + 1) (u + 3))) = 0$

Factor.

Replace all occurrences of $u$ with $x$ .

$2 ((x + 1) (x - 1) ((x + 1) (x + 3))) = 0$

Remove unnecessary parentheses.

$2 ((x + 1) (x - 1) (x + 1) (x + 3)) = 0$

Factor.

Combine exponents.

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

$2 ((x + 1) (x + 1) (x - 1) (x + 3)) = 0$

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

$2 ((x + 1) (x + 1) (x - 1) (x + 3)) = 0$

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

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

Add $1$ and $1$ .

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

Remove unnecessary parentheses.

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

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

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

$x - 1 = 0$

$x + 3 = 0$

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

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

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

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

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = - 1, 1, - 3$

x-intercept(s) in point form.

x-intercept(s): $(- 1, 0), (1, 0), (- 3, 0)$

Step 2

Find the y-intercepts.

To find the y-intercept(s), substitute in $0$ for $x$ and solve for $y$ .

$y = 2 {(0)}^{4} + 8 {(0)}^{3} + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Solve the equation.

Remove parentheses.

$y = 2 \cdot 0^{4} + 8 {(0)}^{3} + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Remove parentheses.

$y = 2 \cdot 0^{4} + 8 \cdot 0^{3} + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Remove parentheses.

$y = 2 \cdot 0^{4} + 8 \cdot 0^{3} + 4 \cdot 0^{2} - 8 \cdot 0 - 6$

Remove parentheses.

$y = 2 {(0)}^{4} + 8 {(0)}^{3} + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Simplify $2 {(0)}^{4} + 8 {(0)}^{3} + 4 {(0)}^{2} - 8 \cdot 0 - 6$ .

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$y = 2 \cdot 0 + 8 {(0)}^{3} + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Multiply $2$ by $0$ .

$y = 0 + 8 {(0)}^{3} + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Raising $0$ to any positive power yields $0$ .

$y = 0 + 8 \cdot 0 + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Multiply $8$ by $0$ .

$y = 0 + 0 + 4 {(0)}^{2} - 8 \cdot 0 - 6$

Raising $0$ to any positive power yields $0$ .

$y = 0 + 0 + 4 \cdot 0 - 8 \cdot 0 - 6$

Multiply $4$ by $0$ .

$y = 0 + 0 + 0 - 8 \cdot 0 - 6$

Multiply $- 8$ by $0$ .

$y = 0 + 0 + 0 + 0 - 6$

Simplify by adding and subtracting.

Add $0$ and $0$ .

$y = 0 + 0 + 0 - 6$

Add $0$ and $0$ .

$y = 0 + 0 - 6$

Add $0$ and $0$ .

$y = 0 - 6$

Subtract $6$ from $0$ .

$y = - 6$

y-intercept(s) in point form.

y-intercept(s): $(0, - 6)$

Step 3

List the intersections.

x-intercept(s): $(- 1, 0), (1, 0), (- 3, 0)$

y-intercept(s): $(0, - 6)$

Step 4

Find the x and y Intercepts y=2x^4+8x^3+4x^2-8x-6

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