2x^4-7x^3+12x^2-8x-8=0 — step-by-step solution

1) Let's solve in

ℝ

2 x^{4}

- 7 x^{3}

+ 12 x^{2}

- 8 x

- 8 = 0

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

x = 2

⇥Let's evaluate the value of:

2 \cdot 2^{4}

- 7 \cdot 2^{3}

+ 12 \cdot 2^{2}

- 8 \cdot 2

- 8

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

2^{4}

2 \cdot 2 \cdot 2 \cdot 2

16

2^{3}

2 \cdot 2 \cdot 2

8

2^{2}

2 \cdot 2

4

■Multiplications are done before additions and subtractions:

8 \cdot 2

16

⇥

2 \cdot 16

- 7 \cdot 8

+ 12 \cdot 4

- 16

- 8

⇥ⓘ Multiplications are done before additions and subtractions:

2 \cdot 16

2 \cdot (10 + 6)

2 \cdot 10 + 2 \cdot 6

20 + 12

32

7 \cdot 8

56

12 \cdot 4

(10 + 2) \cdot 4

10 \cdot 4 + 2 \cdot 4

40 + 8

48

■Start from

-16

and move back

8

more on the number line, away from zero:

- 16 - 8

-24

⇥

32

- 56

+ 48

- 24

⇥ⓘ Start from

32

and move back

56

on the number line, going below zero:

32 - 56

-24

⇥

- 24 + 48 - 24

⇥ⓘ Start from

-24

and move forward

48

on the number line, going past zero:

- 24 + 48

24

⇥

24 - 24

⇥ⓘ Start from

24

and move back

24

on the number line, down to zero:

24 - 24

0

⇥

0

⇥Solution found:

x = 2

⇥1.2) Let's start the euclidean division of

(2 x^{4}

- 7 x^{3}

+ 12 x^{2}

- 8 x

- 8)

(x - 2)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$2 x^{4}$ $- 7 x^{3}$ $+ 12 x^{2}$ $- 8 x$ $- 8$	$\frac{2 x^{4}}{x}$	$2 x^{3}$	$2 x^{3} \cdot (x - 2) = 2 x^{4} - 4 x^{3}$	$(2 x^{4}$ $- 7 x^{3}$ $+ 12 x^{2}$ $- 8 x$ $- 8) - (2 x^{4} - 4 x^{3}) = - 3 x^{3}$ $+ 12 x^{2}$ $- 8 x$ $- 8$
$- 3 x^{3}$ $+ 12 x^{2}$ $- 8 x$ $- 8$	$- \frac{3 x^{3}}{x}$	$- 3 x^{2}$	$- 3 x^{2} \cdot (x - 2) = - 3 x^{3} + 6 x^{2}$	$(- 3 x^{3}$ $+ 12 x^{2}$ $- 8 x$ $- 8) - (- 3 x^{3} + 6 x^{2}) = 6 x^{2} - 8 x - 8$
$6 x^{2} - 8 x - 8$	$\frac{6 x^{2}}{x}$	$6 x$	$6 x \cdot (x - 2) = 6 x^{2} - 12 x$	$(6 x^{2} - 8 x - 8) - (6 x^{2} - 12 x) = 4 x - 8$
$4 x - 8$	$\frac{4 x}{x}$	$4$	$4 \cdot (x - 2) = 4 x - 8$	$(4 x - 8) - (4 x - 8) = 0$

⇥Therefore, the result of the euclidean division is:

2 x^{3}

- 3 x^{2}

+ 6 x

+ 4

⇥1.3) Let's solve in

ℝ

2 x^{3}

- 3 x^{2}

+ 6 x

+ 4 = 0

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

x = - \frac{1}{2}

⇥⇥Let's evaluate the value of:

2 \cdot {(- \frac{1}{2})}^{3}

- 3 \cdot {(- \frac{1}{2})}^{2}

+ 6 \cdot (- \frac{1}{2})

+ 4

⇥⇥ⓘ The minus sign is placed in front of the product:

6 \cdot (- \frac{1}{2})

- 6 \cdot \frac{1}{2}

⇥⇥

2 \cdot {(- \frac{1}{2})}^{3}

- 3 \cdot {(- \frac{1}{2})}^{2}

- 6 \cdot \frac{1}{2}

+ 4

⇥⇥ⓘ Multiplying a fraction by a number multiplies its numerator:

6 \cdot \frac{1}{2}

\frac{6 \cdot 1}{2}

\frac{6}{2}

■We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{6}{2}

-3

⇥⇥

2 \cdot {(- \frac{1}{2})}^{3}

- 3 \cdot {(- \frac{1}{2})}^{2}

- 3

+ 4

⇥⇥ⓘ Start from

-3

and move forward

4

on the number line, going past zero:

- 3 + 4

1

⇥⇥

2 \cdot {(- \frac{1}{2})}^{3} - 3 \cdot {(- \frac{1}{2})}^{2} + 1

⇥⇥ⓘ The power of a quotient is the quotient of the powers, (a/b)^n = a^n/b^n:

{(- \frac{1}{2})}^{3}

\frac{{(-1)}^{3}}{2^{3}}

{(- \frac{1}{2})}^{2}

\frac{{(-1)}^{2}}{2^{2}}

⇥⇥

2 \cdot \frac{{(-1)}^{3}}{2^{3}} - 3 \cdot \frac{{(-1)}^{2}}{2^{2}} + 1

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

{(-1)}^{3}

(-1) \cdot (-1) \cdot (-1)

-1

2^{3}

2 \cdot 2 \cdot 2

8

{(-1)}^{2}

(-1) \cdot (-1)

1

2^{2}

2 \cdot 2

4

⇥⇥

2 \cdot (- \frac{1}{8}) - 3 \cdot \frac{1}{4} + 1

⇥⇥ⓘ The minus sign is placed in front of the product:

2 \cdot (- \frac{1}{8})

- 2 \cdot \frac{1}{8}

■Multiplying a fraction by a number multiplies its numerator:

3 \cdot \frac{1}{4}

\frac{3 \cdot 1}{4}

\frac{3}{4}

■To add fractions, put them over the same denominator, then add the numerators:

- \frac{3}{4} + 1

\frac{- 3 + 4}{4}

⇥⇥

- 2 \cdot \frac{1}{8} + \frac{- 3 + 4}{4}

⇥⇥ⓘ Multiplying a fraction by a number multiplies its numerator:

2 \cdot \frac{1}{8}

\frac{2 \cdot 1}{8}

\frac{2}{8}

■We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{2}{8}

- \frac{1 \cdot 2}{4 \cdot 2}

- \frac{1}{4}

■The numerator and the denominator are computed before the division:

- 3 + 4

1

⇥⇥

- \frac{1}{4} + \frac{1}{4}

⇥⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

- \frac{1}{4} + \frac{1}{4}

\frac{- 1 + 1}{4}

⇥⇥

\frac{- 1 + 1}{4}

⇥⇥ⓘ The numerator and the denominator are computed before the division:

- 1 + 1

0

⇥⇥

\frac{0}{4}

⇥⇥

0

⇥⇥Solution obtained:

x = - \frac{1}{2}

⇥⇥1.3.2) Let's start the euclidean division of

(2 x^{3}

- 3 x^{2}

+ 6 x

+ 4)

(x - (\frac{-1}{2}))

⇥⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$2 x^{3}$ $- 3 x^{2}$ $+ 6 x$ $+ 4$	$\frac{2 x^{3}}{x}$	$2 x^{2}$	$2 x^{2} \cdot (x - (\frac{-1}{2})) = 2 x^{3} + x^{2}$	$(2 x^{3}$ $- 3 x^{2}$ $+ 6 x$ $+ 4) - (2 x^{3} + x^{2}) = - 4 x^{2} + 6 x + 4$
$- 4 x^{2} + 6 x + 4$	$- \frac{4 x^{2}}{x}$	$- 4 x$	$- 4 x \cdot (x - (\frac{-1}{2})) = - 4 x^{2} - 2 x$	$(- 4 x^{2} + 6 x + 4) - (- 4 x^{2} - 2 x) = 8 x + 4$
$8 x + 4$	$\frac{8 x}{x}$	$8$	$8 \cdot (x - (\frac{-1}{2})) = 8 x + 4$	$(8 x + 4) - (8 x + 4) = 0$

⇥⇥So the result of the euclidean division is:

2 x^{2} - 4 x + 8

⇥⇥1.3.3) Let's solve in

ℝ

2 x^{2} - 4 x + 8 = 0

⇥⇥Since

2 \neq 0

, divide both sides by

2

x^{2} - 2 x + 4 = 0

⇥⇥Given the quadratic form

a x^{2} + b x + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -2

and

c = 4

⇥⇥

Δ = {(-2)}^{2} - 4 \cdot 1 \cdot 4

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

{(-2)}^{2}

(-2) \cdot (-2)

4

⇥⇥

Δ = 4 - 4 \cdot 4

⇥⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 4

16

⇥⇥

Δ = 4 - 16

⇥⇥ⓘ Start from

4

and move back

16

on the number line, going below zero:

4 - 16

-12

⇥⇥

Δ = -12

⇥⇥As

Δ < 0

, the equation has no real root.

⇥⇥Solution:

x \in \emptyset

⇥Solution found:

x = - \frac{1}{2}

Solutions found:

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

▶Solution:

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

2) The numeric value of:

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

is:

▷

x = {-0.5, 2}