factor -a^3+4a^2-5a+2 — step-by-step solution

1) Let's factorize in

ℝ

- a^{3}

+ 4 a^{2}

- 5 a

+ 2

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

a = 1

⇥Let's evaluate the value of:

- 1^{3}

+ 4 \cdot 1^{2}

- 5 \cdot 1

+ 2

⇥

- 1

+ 4

- 5

+ 2

⇥ⓘ Start from

-1

and move forward

4

on the number line, going past zero:

- 1 + 4

3

⇥

3 - 5 + 2

⇥ⓘ Start from

3

and move back

5

on the number line, going below zero:

3 - 5

-2

⇥

- 2 + 2

⇥ⓘ Start from

-2

and move forward

2

on the number line, up to zero:

- 2 + 2

0

⇥

0

⇥Root found:

a = 1

⇥1.2) Let's start the euclidean division of

(- a^{3}

+ 4 a^{2}

- 5 a

+ 2)

(a - 1)

⇥

Current expression	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$- a^{3}$ $+ 4 a^{2}$ $- 5 a$ $+ 2$	$- \frac{a^{3}}{a}$	$- a^{2}$	$- a^{2} \cdot (a - 1) = - a^{3} + a^{2}$	$(- a^{3}$ $+ 4 a^{2}$ $- 5 a$ $+ 2) - (- a^{3} + a^{2}) = 3 a^{2} - 5 a + 2$
$3 a^{2} - 5 a + 2$	$\frac{3 a^{2}}{a}$	$3 a$	$3 a \cdot (a - 1) = 3 a^{2} - 3 a$	$(3 a^{2} - 5 a + 2) - (3 a^{2} - 3 a) = - 2 a + 2$
$- 2 a + 2$	$- \frac{2 a}{a}$	$-2$	$-2 \cdot (a - 1) = - 2 a + 2$	$(- 2 a + 2) - (- 2 a + 2) = 0$

⇥Therefore, the result of the euclidean division is:

- a^{2} + 3 a - 2

⇥1.3) Let's factorize in

ℝ

(- a^{2} + 3 a - 2)

⇥Given the quadratic form

A a^{2} + B a + C

, you can calculate the discriminant:

Δ = B^{2} - 4 A C

. Here

A = -1

B = 3

and

C = -2

⇥

Δ = 3^{2} - 4 \cdot (-1) \cdot (-2)

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

3^{2}

3 \cdot 3

9

■Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

-4 \cdot (-1)

4

■The minus sign is placed in front of the product:

4 \cdot (-2)

- 4 \cdot 2

⇥

Δ = 9 - 4 \cdot 2

⇥ⓘ Multiplications are done before additions and subtractions:

4 \cdot 2

8

⇥

Δ = 9 - 8

⇥ⓘ Start from

9

and move back

8

on the number line:

9 - 8

1

⇥

Δ = 1

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

a_{1}

\frac{- B - \sqrt{Δ}}{2 A}

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

⇥

a_{1} = \frac{- 3 - 1}{2 \cdot (-1)}

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

- 3 - 1

-4

■A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-1)

-2

⇥

a_{1} = \frac{-4}{-2}

⇥ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{-4}{-2}

\frac{4}{2}

⇥

a_{1} = \frac{4}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{4}{2}

2

⇥

a_{1} = 2

⇥The quadratic formula gives

a_{2}

\frac{- B + \sqrt{Δ}}{2 A}

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

⇥

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

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

- 3 + 1

-2

■A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

2 \cdot (-1)

-2

⇥

a_{2} = \frac{-2}{-2}

⇥ⓘ The minus signs of the numerator and the denominator cancel out:

\frac{-2}{-2}

\frac{2}{2}

⇥

a_{2} = \frac{2}{2}

⇥ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

⇥

a_{2} = 1

▷The factor

(- a^{2} + 3 a - 2)

factors as:

- (a - 1) \cdot (a - 2)

2) Let's replace each factor by its factorization:

(a - 1) \cdot (- (a - 1) \cdot (a - 2))

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

(a - 1) \cdot (- (a - 1)) \cdot (a - 2)

- (a - 1) \cdot (a - 1) \cdot (a - 2)

ⅰ

- a^{2} + 3 a - 2

- (a - 1) \cdot (a - 2)

- (a - 1) \cdot (a - 1) \cdot (a - 2)

▶Factored expression:

- {(a - 1)}^{2} \cdot (a - 2)