Algebra Examples

$y = x^{4} - 20 x^{3} + 148 x^{2} - 480 x + 576$

Step 1

Write $y = x^{4} - 20 x^{3} + 148 x^{2} - 480 x + 576$ as a function.

$f (x) = x^{4} - 20 x^{3} + 148 x^{2} - 480 x + 576$

Step 2

Find the first derivative of the function.

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Differentiate.

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By the Sum Rule, the derivative of $x^{4} - 20 x^{3} + 148 x^{2} - 480 x + 576$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$ .

$\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$4 x^{3} + \frac{d}{d x} [- 20 x^{3}] + \frac{d}{d x} [148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Evaluate $\frac{d}{d x} [- 20 x^{3}]$ .

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Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x^{3}$ with respect to $x$ is $- 20 \frac{d}{d x} [x^{3}]$ .

$4 x^{3} - 20 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$4 x^{3} - 20 (3 x^{2}) + \frac{d}{d x} [148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Multiply $3$ by $- 20$ .

$4 x^{3} - 60 x^{2} + \frac{d}{d x} [148 x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Evaluate $\frac{d}{d x} [148 x^{2}]$ .

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Since $148$ is constant with respect to $x$ , the derivative of $148 x^{2}$ with respect to $x$ is $148 \frac{d}{d x} [x^{2}]$ .

$4 x^{3} - 60 x^{2} + 148 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$4 x^{3} - 60 x^{2} + 148 (2 x) + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Multiply $2$ by $148$ .

$4 x^{3} - 60 x^{2} + 296 x + \frac{d}{d x} [- 480 x] + \frac{d}{d x} [576]$

Evaluate $\frac{d}{d x} [- 480 x]$ .

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Since $- 480$ is constant with respect to $x$ , the derivative of $- 480 x$ with respect to $x$ is $- 480 \frac{d}{d x} [x]$ .

$4 x^{3} - 60 x^{2} + 296 x - 480 \frac{d}{d x} [x] + \frac{d}{d x} [576]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$4 x^{3} - 60 x^{2} + 296 x - 480 \cdot 1 + \frac{d}{d x} [576]$

Multiply $- 480$ by $1$ .

$4 x^{3} - 60 x^{2} + 296 x - 480 + \frac{d}{d x} [576]$

Differentiate using the Constant Rule.

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Since $576$ is constant with respect to $x$ , the derivative of $576$ with respect to $x$ is $0$ .

$4 x^{3} - 60 x^{2} + 296 x - 480 + 0$

Add $4 x^{3} - 60 x^{2} + 296 x - 480$ and $0$ .

$4 x^{3} - 60 x^{2} + 296 x - 480$

Step 3

Find the second derivative of the function.

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By the Sum Rule, the derivative of $4 x^{3} - 60 x^{2} + 296 x - 480$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [- 60 x^{2}] + \frac{d}{d x} [296 x] + \frac{d}{d x} [- 480]$ .

$f'' (x) = \frac{d}{d x} (4 x^{3}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (296 x) + \frac{d}{d x} (- 480)$

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

$f'' (x) = 4 \frac{d}{d x} (x^{3}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (296 x) + \frac{d}{d x} (- 480)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f'' (x) = 4 (3 x^{2}) + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (296 x) + \frac{d}{d x} (- 480)$

Multiply $3$ by $4$ .

$f'' (x) = 12 x^{2} + \frac{d}{d x} (- 60 x^{2}) + \frac{d}{d x} (296 x) + \frac{d}{d x} (- 480)$

Evaluate $\frac{d}{d x} [- 60 x^{2}]$ .

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Since $- 60$ is constant with respect to $x$ , the derivative of $- 60 x^{2}$ with respect to $x$ is $- 60 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 12 x^{2} - 60 \frac{d}{d x} x^{2} + \frac{d}{d x} (296 x) + \frac{d}{d x} (- 480)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 12 x^{2} - 60 (2 x) + \frac{d}{d x} (296 x) + \frac{d}{d x} (- 480)$

Multiply $2$ by $- 60$ .

$f'' (x) = 12 x^{2} - 120 x + \frac{d}{d x} (296 x) + \frac{d}{d x} (- 480)$

Evaluate $\frac{d}{d x} [296 x]$ .

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Since $296$ is constant with respect to $x$ , the derivative of $296 x$ with respect to $x$ is $296 \frac{d}{d x} [x]$ .

$f'' (x) = 12 x^{2} - 120 x + 296 \frac{d}{d x} (x) + \frac{d}{d x} (- 480)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 12 x^{2} - 120 x + 296 \cdot 1 + \frac{d}{d x} (- 480)$

Multiply $296$ by $1$ .

$f'' (x) = 12 x^{2} - 120 x + 296 + \frac{d}{d x} (- 480)$

Differentiate using the Constant Rule.

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Since $- 480$ is constant with respect to $x$ , the derivative of $- 480$ with respect to $x$ is $0$ .

$f'' (x) = 12 x^{2} - 120 x + 296 + 0$

Add $12 x^{2} - 120 x + 296$ and $0$ .

$f'' (x) = 12 x^{2} - 120 x + 296$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$4 x^{3} - 60 x^{2} + 296 x - 480 = 0$

Step 5

Find the first derivative.

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Find the first derivative.

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Differentiate.

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$f' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f' (x) = 4 x^{3} + \frac{d}{d x} (- 20 x^{3}) + \frac{d}{d x} (148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Evaluate $\frac{d}{d x} [- 20 x^{3}]$ .

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Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x^{3}$ with respect to $x$ is $- 20 \frac{d}{d x} [x^{3}]$ .

$f' (x) = 4 x^{3} - 20 \frac{d}{d x} x^{3} + \frac{d}{d x} (148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$f' (x) = 4 x^{3} - 20 (3 x^{2}) + \frac{d}{d x} (148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Multiply $3$ by $- 20$ .

$f' (x) = 4 x^{3} - 60 x^{2} + \frac{d}{d x} (148 x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Evaluate $\frac{d}{d x} [148 x^{2}]$ .

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Since $148$ is constant with respect to $x$ , the derivative of $148 x^{2}$ with respect to $x$ is $148 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 148 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 148 (2 x) + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Multiply $2$ by $148$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 296 x + \frac{d}{d x} (- 480 x) + \frac{d}{d x} (576)$

Evaluate $\frac{d}{d x} [- 480 x]$ .

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Since $- 480$ is constant with respect to $x$ , the derivative of $- 480 x$ with respect to $x$ is $- 480 \frac{d}{d x} [x]$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 296 x - 480 \frac{d}{d x} x + \frac{d}{d x} (576)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 296 x - 480 \cdot 1 + \frac{d}{d x} (576)$

Multiply $- 480$ by $1$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 296 x - 480 + \frac{d}{d x} (576)$

Differentiate using the Constant Rule.

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Since $576$ is constant with respect to $x$ , the derivative of $576$ with respect to $x$ is $0$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 296 x - 480 + 0$

Add $4 x^{3} - 60 x^{2} + 296 x - 480$ and $0$ .

$f' (x) = 4 x^{3} - 60 x^{2} + 296 x - 480$

The first derivative of $f (x)$ with respect to $x$ is $4 x^{3} - 60 x^{2} + 296 x - 480$ .

$4 x^{3} - 60 x^{2} + 296 x - 480$

Step 6

Set the first derivative equal to $0$ then solve the equation $4 x^{3} - 60 x^{2} + 296 x - 480 = 0$ .

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Set the first derivative equal to $0$ .

$4 x^{3} - 60 x^{2} + 296 x - 480 = 0$

Factor the left side of the equation.

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Factor $4$ out of $4 x^{3} - 60 x^{2} + 296 x - 480$ .

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

$4 (x^{3}) - 60 x^{2} + 296 x - 480 = 0$

Factor $4$ out of $- 60 x^{2}$ .

$4 (x^{3}) + 4 (- 15 x^{2}) + 296 x - 480 = 0$

Factor $4$ out of $296 x$ .

$4 (x^{3}) + 4 (- 15 x^{2}) + 4 (74 x) - 480 = 0$

Factor $4$ out of $- 480$ .

$4 x^{3} + 4 (- 15 x^{2}) + 4 (74 x) + 4 \cdot - 120 = 0$

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

$4 (x^{3} - 15 x^{2}) + 4 (74 x) + 4 \cdot - 120 = 0$

Factor $4$ out of $4 (x^{3} - 15 x^{2}) + 4 (74 x)$ .

$4 (x^{3} - 15 x^{2} + 74 x) + 4 \cdot - 120 = 0$

Factor $4$ out of $4 (x^{3} - 15 x^{2} + 74 x) + 4 \cdot - 120$ .

$4 (x^{3} - 15 x^{2} + 74 x - 120) = 0$

Factor $x^{3} - 15 x^{2} + 74 x - 120$ using the rational roots test.

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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.

$p = \pm 1, \pm 120, \pm 2, \pm 60, \pm 3, \pm 40, \pm 4, \pm 30, \pm 5, \pm 24, \pm 6, \pm 20, \pm 8, \pm 15, \pm 10, \pm 12 q = \pm 1$

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

$\pm 1, \pm 120, \pm 2, \pm 60, \pm 3, \pm 40, \pm 4, \pm 30, \pm 5, \pm 24, \pm 6, \pm 20, \pm 8, \pm 15, \pm 10, \pm 12$

Substitute $4$ and simplify the expression. In this case, the expression is equal to $0$ so $4$ is a root of the polynomial.

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Substitute $4$ into the polynomial.

$4^{3} - 15 \cdot 4^{2} + 74 \cdot 4 - 120$

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

$64 - 15 \cdot 4^{2} + 74 \cdot 4 - 120$

Raise $4$ to the power of $2$ .

$64 - 15 \cdot 16 + 74 \cdot 4 - 120$

Multiply $- 15$ by $16$ .

$64 - 240 + 74 \cdot 4 - 120$

Subtract $240$ from $64$ .

$- 176 + 74 \cdot 4 - 120$

Multiply $74$ by $4$ .

$- 176 + 296 - 120$

Add $- 176$ and $296$ .

$120 - 120$

Subtract $120$ from $120$ .

$0$

Since $4$ is a known root, divide the polynomial by $x - 4$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 15 x^{2} + 74 x - 120}{x - 4}$

Divide $x^{3} - 15 x^{2} + 74 x - 120$ by $x - 4$ .

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

$x$

$4$

$x^{3}$

$15 x^{2}$

$74 x$

$120$

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

					$x^{2}$
	$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			+	$x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} - 4 x^{2}$

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$

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

				$x^{2}$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$

Divide the highest order term in the dividend $- 11 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$11 x$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$11 x$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					-	$11 x^{2}$	+	$44 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 11 x^{2} + 44 x$

				$x^{2}$	-	$11 x$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					+	$11 x^{2}$	-	$44 x$

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

				$x^{2}$	-	$11 x$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					+	$11 x^{2}$	-	$44 x$
							+	$30 x$

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

				$x^{2}$	-	$11 x$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					+	$11 x^{2}$	-	$44 x$
							+	$30 x$	-	$120$

Divide the highest order term in the dividend $30 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$11 x$	+	$30$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					+	$11 x^{2}$	-	$44 x$
							+	$30 x$	-	$120$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$11 x$	+	$30$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					+	$11 x^{2}$	-	$44 x$
							+	$30 x$	-	$120$
							+	$30 x$	-	$120$

The expression needs to be subtracted from the dividend, so change all the signs in $30 x - 120$

				$x^{2}$	-	$11 x$	+	$30$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					+	$11 x^{2}$	-	$44 x$
							+	$30 x$	-	$120$
							-	$30 x$	+	$120$

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

				$x^{2}$	-	$11 x$	+	$30$
$x$	-	$4$		$x^{3}$	-	$15 x^{2}$	+	$74 x$	-	$120$
			-	$x^{3}$	+	$4 x^{2}$
					-	$11 x^{2}$	+	$74 x$
					+	$11 x^{2}$	-	$44 x$
							+	$30 x$	-	$120$
							-	$30 x$	+	$120$
										$0$

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

$x^{2} - 11 x + 30$

Write $x^{3} - 15 x^{2} + 74 x - 120$ as a set of factors.

$4 ((x - 4) (x^{2} - 11 x + 30)) = 0$

Factor.

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Factor $x^{2} - 11 x + 30$ using the AC method.

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Factor $x^{2} - 11 x + 30$ using the AC method.

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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 $30$ and whose sum is $- 11$ .

$- 6, - 5$

Write the factored form using these integers.

$4 ((x - 4) ((x - 6) (x - 5))) = 0$

Remove unnecessary parentheses.

$4 ((x - 4) (x - 6) (x - 5)) = 0$

Remove unnecessary parentheses.

$4 (x - 4) (x - 6) (x - 5) = 0$

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

$x - 4 = 0$

$x - 6 = 0$

$x - 5 = 0$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x - 6$ equal to $0$ and solve for $x$ .

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Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x - 5$ equal to $0$ and solve for $x$ .

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Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

The final solution is all the values that make $4 (x - 4) (x - 6) (x - 5) = 0$ true.

$x = 4, 6, 5$

Step 7

Find the values where the derivative is undefined.

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The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 8

Critical points to evaluate.

$x = 4, 6, 5$

Step 9

Evaluate the second derivative at $x = 4$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(4)}^{2} - 120 \cdot 4 + 296$

Step 10

Evaluate the second derivative.

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Simplify each term.

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Raise $4$ to the power of $2$ .

$12 \cdot 16 - 120 \cdot 4 + 296$

Multiply $12$ by $16$ .

$192 - 120 \cdot 4 + 296$

Multiply $- 120$ by $4$ .

$192 - 480 + 296$

Simplify by adding and subtracting.

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Subtract $480$ from $192$ .

$- 288 + 296$

Add $- 288$ and $296$ .

$8$

Step 11

$x = 4$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 4$ is a local minimum

Step 12

Find the y-value when $x = 4$ .

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Replace the variable $x$ with $4$ in the expression.

$f (4) = {(4)}^{4} - 20 {(4)}^{3} + 148 {(4)}^{2} - 480 \cdot 4 + 576$

Simplify the result.

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Simplify each term.

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

$f (4) = 256 - 20 {(4)}^{3} + 148 {(4)}^{2} - 480 \cdot 4 + 576$

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

$f (4) = 256 - 20 \cdot 64 + 148 {(4)}^{2} - 480 \cdot 4 + 576$

Multiply $- 20$ by $64$ .

$f (4) = 256 - 1280 + 148 {(4)}^{2} - 480 \cdot 4 + 576$

Raise $4$ to the power of $2$ .

$f (4) = 256 - 1280 + 148 \cdot 16 - 480 \cdot 4 + 576$

Multiply $148$ by $16$ .

$f (4) = 256 - 1280 + 2368 - 480 \cdot 4 + 576$

Multiply $- 480$ by $4$ .

$f (4) = 256 - 1280 + 2368 - 1920 + 576$

Simplify by adding and subtracting.

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Subtract $1280$ from $256$ .

$f (4) = - 1024 + 2368 - 1920 + 576$

Add $- 1024$ and $2368$ .

$f (4) = 1344 - 1920 + 576$

Subtract $1920$ from $1344$ .

$f (4) = - 576 + 576$

Add $- 576$ and $576$ .

$f (4) = 0$

The final answer is $0$ .

$y = 0$

Step 13

Evaluate the second derivative at $x = 6$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(6)}^{2} - 120 \cdot 6 + 296$

Step 14

Evaluate the second derivative.

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Simplify each term.

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Raise $6$ to the power of $2$ .

$12 \cdot 36 - 120 \cdot 6 + 296$

Multiply $12$ by $36$ .

$432 - 120 \cdot 6 + 296$

Multiply $- 120$ by $6$ .

$432 - 720 + 296$

Simplify by adding and subtracting.

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Subtract $720$ from $432$ .

$- 288 + 296$

Add $- 288$ and $296$ .

$8$

Step 15

$x = 6$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 6$ is a local minimum

Step 16

Find the y-value when $x = 6$ .

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Replace the variable $x$ with $6$ in the expression.

$f (6) = {(6)}^{4} - 20 {(6)}^{3} + 148 {(6)}^{2} - 480 \cdot 6 + 576$

Simplify the result.

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Simplify each term.

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Raise $6$ to the power of $4$ .

$f (6) = 1296 - 20 {(6)}^{3} + 148 {(6)}^{2} - 480 \cdot 6 + 576$

Raise $6$ to the power of $3$ .

$f (6) = 1296 - 20 \cdot 216 + 148 {(6)}^{2} - 480 \cdot 6 + 576$

Multiply $- 20$ by $216$ .

$f (6) = 1296 - 4320 + 148 {(6)}^{2} - 480 \cdot 6 + 576$

Raise $6$ to the power of $2$ .

$f (6) = 1296 - 4320 + 148 \cdot 36 - 480 \cdot 6 + 576$

Multiply $148$ by $36$ .

$f (6) = 1296 - 4320 + 5328 - 480 \cdot 6 + 576$

Multiply $- 480$ by $6$ .

$f (6) = 1296 - 4320 + 5328 - 2880 + 576$

Simplify by adding and subtracting.

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Subtract $4320$ from $1296$ .

$f (6) = - 3024 + 5328 - 2880 + 576$

Add $- 3024$ and $5328$ .

$f (6) = 2304 - 2880 + 576$

Subtract $2880$ from $2304$ .

$f (6) = - 576 + 576$

Add $- 576$ and $576$ .

$f (6) = 0$

The final answer is $0$ .

$y = 0$

Step 17

Evaluate the second derivative at $x = 5$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 {(5)}^{2} - 120 \cdot 5 + 296$

Step 18

Evaluate the second derivative.

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Simplify each term.

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Raise $5$ to the power of $2$ .

$12 \cdot 25 - 120 \cdot 5 + 296$

Multiply $12$ by $25$ .

$300 - 120 \cdot 5 + 296$

Multiply $- 120$ by $5$ .

$300 - 600 + 296$

Simplify by adding and subtracting.

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Subtract $600$ from $300$ .

$- 300 + 296$

Add $- 300$ and $296$ .

$- 4$

Step 19

$x = 5$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 5$ is a local maximum

Step 20

Find the y-value when $x = 5$ .

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Replace the variable $x$ with $5$ in the expression.

$f (5) = {(5)}^{4} - 20 {(5)}^{3} + 148 {(5)}^{2} - 480 \cdot 5 + 576$

Simplify the result.

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Simplify each term.

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Raise $5$ to the power of $4$ .

$f (5) = 625 - 20 {(5)}^{3} + 148 {(5)}^{2} - 480 \cdot 5 + 576$

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

$f (5) = 625 - 20 \cdot 125 + 148 {(5)}^{2} - 480 \cdot 5 + 576$

Multiply $- 20$ by $125$ .

$f (5) = 625 - 2500 + 148 {(5)}^{2} - 480 \cdot 5 + 576$

Raise $5$ to the power of $2$ .

$f (5) = 625 - 2500 + 148 \cdot 25 - 480 \cdot 5 + 576$

Multiply $148$ by $25$ .

$f (5) = 625 - 2500 + 3700 - 480 \cdot 5 + 576$

Multiply $- 480$ by $5$ .

$f (5) = 625 - 2500 + 3700 - 2400 + 576$

Simplify by adding and subtracting.

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Subtract $2500$ from $625$ .

$f (5) = - 1875 + 3700 - 2400 + 576$

Add $- 1875$ and $3700$ .

$f (5) = 1825 - 2400 + 576$

Subtract $2400$ from $1825$ .

$f (5) = - 575 + 576$

Add $- 575$ and $576$ .

$f (5) = 1$

The final answer is $1$ .

$y = 1$

Step 21

These are the local extrema for $f (x) = x^{4} - 20 x^{3} + 148 x^{2} - 480 x + 576$ .

$(4, 0)$ is a local minima

$(6, 0)$ is a local minima

$(5, 1)$ is a local maxima

Step 22