x^4+4x^3-4x^2+4x-5<0 — step-by-step solution

1) Let's solve in

ℝ

x^{4}

+ 4 x^{3}

- 4 x^{2}

+ 4 x

- 5 < 0

⇥1.1) Try the obvious solution first:

x = 1

⇥Let's evaluate the value of:

1^{4}

+ 4 \cdot 1^{3}

- 4 \cdot 1^{2}

+ 4 \cdot 1

- 5

⇥

1

+ 4

- 4

+ 4

- 5

⇥ⓘ Start from

4

and move back

4

on the number line, down to zero:

4 - 4

0

⇥

1

+ 0

+ 4

- 5

⇥ⓘ Adding 0 changes nothing:

0 + 4

4

■Start from

1

and move forward

4

on the number line:

1 + 4

5

⇥

5 - 5

⇥ⓘ Start from

5

and move back

5

on the number line, down to zero:

5 - 5

0

⇥

0

⇥Solution found:

x = 1

⇥1.2) Compute the quotient and remainder of

(x^{4}

+ 4 x^{3}

- 4 x^{2}

+ 4 x

- 5)

(x - 1)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{4}$ $+ 4 x^{3}$ $- 4 x^{2}$ $+ 4 x$ $- 5$	$\frac{x^{4}}{x}$	$x^{3}$	$x^{3} \cdot (x - 1) = x^{4} - x^{3}$	$(x^{4}$ $+ 4 x^{3}$ $- 4 x^{2}$ $+ 4 x$ $- 5) - (x^{4} - x^{3}) = 5 x^{3}$ $- 4 x^{2}$ $+ 4 x$ $- 5$
$5 x^{3}$ $- 4 x^{2}$ $+ 4 x$ $- 5$	$\frac{5 x^{3}}{x}$	$5 x^{2}$	$5 x^{2} \cdot (x - 1) = 5 x^{3} - 5 x^{2}$	$(5 x^{3}$ $- 4 x^{2}$ $+ 4 x$ $- 5) - (5 x^{3} - 5 x^{2}) = x^{2} + 4 x - 5$
$x^{2} + 4 x - 5$	$\frac{x^{2}}{x}$	$x$	$x \cdot (x - 1) = x^{2} - x$	$(x^{2} + 4 x - 5) - (x^{2} - x) = 5 x - 5$
$5 x - 5$	$\frac{5 x}{x}$	$5$	$5 \cdot (x - 1) = 5 x - 5$	$(5 x - 5) - (5 x - 5) = 0$

⇥Therefore, the result of the euclidean division is:

x^{3}

+ 5 x^{2}

+ x

+ 5

(x - 1) \cdot (x^{3}

+ 5 x^{2}

+ x

+ 5)

We study the sign of each factor of the product

⇥The expression

x - 1

is zero at

x = 1

, it is negative before and positive after

⇥1.3) Let's study the sign of

x^{3}

+ 5 x^{2}

+ x

+ 5

by widening the sign to ≤ and solving in

ℝ

x^{3}

+ 5 x^{2}

+ x

+ 5 \leq 0

⇥⇥1.3.1) Let's start with the obvious candidate:

x = -5

⇥⇥Let's compute the value of:

{(-5)}^{3}

+ 5 \cdot {(-5)}^{2}

- 5

+ 5

⇥⇥ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-5)}^{3}

(-5) \cdot (-5) \cdot (-5)

-125

{(-5)}^{2}

(-5) \cdot (-5)

25

■Start from

-5

and move forward

5

on the number line, up to zero:

- 5 + 5

0

⇥⇥

- 125 + 5 \cdot 25 + 0

⇥⇥ⓘ Adding 0 changes nothing:

5 \cdot 25 + 0

5 \cdot 25

■Multiplications are done before additions and subtractions:

5 \cdot 25

5 \cdot (20 + 5)

5 \cdot 20 + 5 \cdot 5

100 + 25

125

⇥⇥

- 125 + 125

⇥⇥ⓘ Start from

-125

and move forward

125

on the number line, up to zero:

- 125 + 125

0

⇥⇥

0

⇥⇥Solution found:

x = -5

⇥⇥1.3.2) Perform polynomial long division of

(x^{3}

+ 5 x^{2}

+ x

+ 5)

(x + 5)

⇥⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$x^{3}$ $+ 5 x^{2}$ $+ x$ $+ 5$	$\frac{x^{3}}{x}$	$x^{2}$	$x^{2} \cdot (x + 5) = x^{3} + 5 x^{2}$	$(x^{3}$ $+ 5 x^{2}$ $+ x$ $+ 5) - (x^{3} + 5 x^{2}) = x + 5$
$x + 5$	$\frac{x}{x}$	$1$	$x + 5 = x + 5$	$(x + 5) - (x + 5) = 0$

⇥⇥So the result of the euclidean division is:

x^{2} + 1

⇥ⅰ

x^{3}

+ 5 x^{2}

+ x

+ 5

(x + 5) \cdot (x^{2} + 1)

⇥

(x + 5) \cdot (x^{2} + 1)

⇥We study the sign of each factor of the product

⇥⇥The expression

x + 5

is zero at

x = -5

, it is negative before and positive after

⇥⇥1.3.3) Let's study the sign of

x^{2} + 1

by solving in

ℝ

x^{2} + 1 \leq 0

⇥⇥In ℝ, an even power is never negative:

x^{2} \leq -1

has no solution

⇥⇥Solution:

x \in \emptyset

⇥⇥So the expression

x^{2} + 1

is always positive

2) Let's draw up the sign table:

x	$-∞$		$-5$		$1$		$+∞$
$x - 1$		-		-	0	+
$x + 5$		-	0	+		+
$x^{2} + 1$		+		+		+
$x^{4}$ $+ 4 x^{3}$ $- 4 x^{2}$ $+ 4 x$ $- 5$		+	0	-	0	+

▶Solution obtained:

x \in (-5, 1)