Find the x and y Intercepts 7x^4-19x^3-6x^2+28x+8

$7 x^{4} - 19 x^{3} - 6 x^{2} + 28 x + 8$

Step 1

Write $7 x^{4} - 19 x^{3} - 6 x^{2} + 28 x + 8$ as an equation.

$y = 7 x^{4} - 19 x^{3} - 6 x^{2} + 28 x + 8$

Step 2

Find the x-intercepts.

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

$0 = 7 x^{4} - 19 x^{3} - 6 x^{2} + 28 x + 8$

Solve the equation.

Rewrite the equation as $7 x^{4} - 19 x^{3} - 6 x^{2} + 28 x + 8 = 0$ .

$7 x^{4} - 19 x^{3} - 6 x^{2} + 28 x + 8 = 0$

Factor the left side of the equation.

Regroup terms.

$28 x + 8 + 7 x^{4} - 19 x^{3} - 6 x^{2} = 0$

Factor $4$ out of $28 x + 8$ .

Factor $4$ out of $28 x$ .

$4 (7 x) + 8 + 7 x^{4} - 19 x^{3} - 6 x^{2} = 0$

Factor $4$ out of $8$ .

$4 (7 x) + 4 (2) + 7 x^{4} - 19 x^{3} - 6 x^{2} = 0$

Factor $4$ out of $4 (7 x) + 4 (2)$ .

$4 (7 x + 2) + 7 x^{4} - 19 x^{3} - 6 x^{2} = 0$

Factor $x^{2}$ out of $7 x^{4} - 19 x^{3} - 6 x^{2}$ .

Factor $x^{2}$ out of $7 x^{4}$ .

$4 (7 x + 2) + x^{2} (7 x^{2}) - 19 x^{3} - 6 x^{2} = 0$

Factor $x^{2}$ out of $- 19 x^{3}$ .

$4 (7 x + 2) + x^{2} (7 x^{2}) + x^{2} (- 19 x) - 6 x^{2} = 0$

Factor $x^{2}$ out of $- 6 x^{2}$ .

$4 (7 x + 2) + x^{2} (7 x^{2}) + x^{2} (- 19 x) + x^{2} \cdot - 6 = 0$

Factor $x^{2}$ out of $x^{2} (7 x^{2}) + x^{2} (- 19 x)$ .

$4 (7 x + 2) + x^{2} (7 x^{2} - 19 x) + x^{2} \cdot - 6 = 0$

Factor $x^{2}$ out of $x^{2} (7 x^{2} - 19 x) + x^{2} \cdot - 6$ .

$4 (7 x + 2) + x^{2} (7 x^{2} - 19 x - 6) = 0$

Factor.

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 $a \cdot c = 7 \cdot - 6 = - 42$ and whose sum is $b = - 19$ .

Factor $- 19$ out of $- 19 x$ .

$4 (7 x + 2) + x^{2} (7 x^{2} - 19 x - 6) = 0$

Rewrite $- 19$ as $2$ plus $- 21$

$4 (7 x + 2) + x^{2} (7 x^{2} + (2 - 21) x - 6) = 0$

Apply the distributive property.

$4 (7 x + 2) + x^{2} (7 x^{2} + 2 x - 21 x - 6) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$4 (7 x + 2) + x^{2} ((7 x^{2} + 2 x) - 21 x - 6) = 0$

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

$4 (7 x + 2) + x^{2} (x (7 x + 2) - 3 (7 x + 2)) = 0$

Factor the polynomial by factoring out the greatest common factor, $7 x + 2$ .

$4 (7 x + 2) + x^{2} ((7 x + 2) (x - 3)) = 0$

Remove unnecessary parentheses.

$4 (7 x + 2) + x^{2} (7 x + 2) (x - 3) = 0$

Factor $7 x + 2$ out of $4 (7 x + 2) + x^{2} (7 x + 2) (x - 3)$ .

Factor $7 x + 2$ out of $4 (7 x + 2)$ .

$(7 x + 2) \cdot 4 + x^{2} (7 x + 2) (x - 3) = 0$

Factor $7 x + 2$ out of $x^{2} (7 x + 2) (x - 3)$ .

$(7 x + 2) \cdot 4 + (7 x + 2) (x^{2} (x - 3)) = 0$

Factor $7 x + 2$ out of $(7 x + 2) \cdot 4 + (7 x + 2) (x^{2} (x - 3))$ .

$(7 x + 2) (4 + x^{2} (x - 3)) = 0$

Apply the distributive property.

$(7 x + 2) (4 + x^{2} x + x^{2} \cdot - 3) = 0$

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

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

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

$(7 x + 2) (4 + x^{2} x + x^{2} \cdot - 3) = 0$

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

$(7 x + 2) (4 + x^{2 + 1} + x^{2} \cdot - 3) = 0$

Add $2$ and $1$ .

$(7 x + 2) (4 + x^{3} + x^{2} \cdot - 3) = 0$

Move $- 3$ to the left of $x^{2}$ .

$(7 x + 2) (4 + x^{3} - 3 x^{2}) = 0$

Reorder terms.

$(7 x + 2) (x^{3} - 3 x^{2} + 4) = 0$

Factor.

Rewrite $x^{3} - 3 x^{2} + 4$ in a factored form.

Factor $x^{3} - 3 x^{2} + 4$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 4, \pm 2 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 4, \pm 2$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

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

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

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

Raise $- 1$ to the power of $2$ .

$- 1 - 3 \cdot 1 + 4$

Multiply $- 3$ by $1$ .

$- 1 - 3 + 4$

Subtract $3$ from $- 1$ .

$- 4 + 4$

Add $- 4$ and $4$ .

$0$

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 3 x^{2} + 4}{x + 1}$

Divide $x^{3} - 3 x^{2} + 4$ by $x + 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{3}$

$3 x^{2}$

$0 x$

$4$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

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

Multiply the new quotient term by the divisor.

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

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} + x^{2}$

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

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

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

Pull the next terms from the original dividend down into the current dividend.

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

Divide the highest order term in the dividend $- 4 x^{2}$ by the highest order term in divisor $x$ .

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

Multiply the new quotient term by the divisor.

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

The expression needs to be subtracted from the dividend, so change all the signs in $- 4 x^{2} - 4 x$

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

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

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

Pull the next terms from the original dividend down into the current dividend.

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

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

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

Multiply the new quotient term by the divisor.

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

The expression needs to be subtracted from the dividend, so change all the signs in $4 x + 4$

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

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

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

Since the remander is $0$ , the final answer is the quotient.

$x^{2} - 4 x + 4$

Write $x^{3} - 3 x^{2} + 4$ as a set of factors.

$(7 x + 2) ((x + 1) (x^{2} - 4 x + 4)) = 0$

Factor using the perfect square rule.

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

$(7 x + 2) ((x + 1) (x^{2} - 4 x + 2^{2})) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$(7 x + 2) ((x + 1) (x^{2} - 2 \cdot x \cdot 2 + 2^{2})) = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$(7 x + 2) ((x + 1) {(x - 2)}^{2}) = 0$

Remove unnecessary parentheses.

$(7 x + 2) (x + 1) {(x - 2)}^{2} = 0$

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

$7 x + 2 = 0$

$x + 1 = 0$

${(x - 2)}^{2} = 0$

Set $7 x + 2$ equal to $0$ and solve for $x$ .

Set $7 x + 2$ equal to $0$ .

$7 x + 2 = 0$

Solve $7 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$7 x = - 2$

Divide each term in $7 x = - 2$ by $7$ and simplify.

Divide each term in $7 x = - 2$ by $7$ .

$\frac{7 x}{7} = \frac{- 2}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 2}{7}$

Divide $x$ by $1$ .

$x = \frac{- 2}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{7}$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

${(x - 2)}^{2} = 0$

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

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(7 x + 2) (x + 1) {(x - 2)}^{2} = 0$ true.

$x = - \frac{2}{7}, - 1, 2$

x-intercept(s) in point form.

x-intercept(s): $(- \frac{2}{7}, 0), (- 1, 0), (2, 0)$

Step 3

Find the y-intercepts.

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

$y = 7 {(0)}^{4} - 19 {(0)}^{3} - 6 {(0)}^{2} + 28 (0) + 8$

Solve the equation.

Remove parentheses.

$y = 7 \cdot 0^{4} - 19 {(0)}^{3} - 6 {(0)}^{2} + 28 (0) + 8$

Remove parentheses.

$y = 7 \cdot 0^{4} - 19 \cdot 0^{3} - 6 {(0)}^{2} + 28 (0) + 8$

Remove parentheses.

$y = 7 \cdot 0^{4} - 19 \cdot 0^{3} - 6 \cdot 0^{2} + 28 (0) + 8$

Remove parentheses.

$y = 7 {(0)}^{4} - 19 {(0)}^{3} - 6 {(0)}^{2} + 28 (0) + 8$

Simplify $7 {(0)}^{4} - 19 {(0)}^{3} - 6 {(0)}^{2} + 28 (0) + 8$ .

Simplify each term.

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

$y = 7 \cdot 0 - 19 {(0)}^{3} - 6 {(0)}^{2} + 28 (0) + 8$

Multiply $7$ by $0$ .

$y = 0 - 19 {(0)}^{3} - 6 {(0)}^{2} + 28 (0) + 8$

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

$y = 0 - 19 \cdot 0 - 6 {(0)}^{2} + 28 (0) + 8$

Multiply $- 19$ by $0$ .

$y = 0 + 0 - 6 {(0)}^{2} + 28 (0) + 8$

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

$y = 0 + 0 - 6 \cdot 0 + 28 (0) + 8$

Multiply $- 6$ by $0$ .

$y = 0 + 0 + 0 + 28 (0) + 8$

Multiply $28$ by $0$ .

$y = 0 + 0 + 0 + 0 + 8$

Simplify by adding numbers.

Add $0$ and $0$ .

$y = 0 + 0 + 0 + 8$

Add $0$ and $0$ .

$y = 0 + 0 + 8$

Add $0$ and $0$ .

$y = 0 + 8$

Add $0$ and $8$ .

$y = 8$

y-intercept(s) in point form.

y-intercept(s): $(0, 8)$

Step 4

List the intersections.

x-intercept(s): $(- \frac{2}{7}, 0), (- 1, 0), (2, 0)$

y-intercept(s): $(0, 8)$

Step 5

Find the x and y Intercepts 7x^4-19x^3-6x^2+28x+8

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