factor 4k^3+12k^2+11k+3 — step-by-step solution

1) Let's factor in

ℝ

4 k^{3}

+ 12 k^{2}

+ 11 k

+ 3

⇥1.1) Try the obvious solution first:

k = -1

⇥Let's evaluate the value of:

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

+ 12 \cdot {(-1)}^{2}

+ 11 \cdot (-1)

+ 3

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

{(-1)}^{3}

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

-1

{(-1)}^{2}

(-1) \cdot (-1)

1

■Multiplying by -1 gives the opposite:

4 \cdot (-1)

-4

■Multiplying by 1 changes nothing:

12 \cdot 1

12

■Multiplications are done before additions and subtractions. A single minus sign does not cancel out, the product of a negative number by a positive number is negative:

11 \cdot (-1)

-11

⇥

- 4

+ 12

- 11

+ 3

⇥ⓘ Start from

-4

and move forward

12

on the number line, going past zero:

- 4 + 12

8

⇥

8 - 11 + 3

⇥ⓘ Start from

8

and move back

11

on the number line, going below zero:

8 - 11

-3

⇥

- 3 + 3

⇥ⓘ Start from

-3

and move forward

3

on the number line, up to zero:

- 3 + 3

0

⇥

0

⇥Root found:

k = -1

⇥1.2) Compute the quotient and remainder of

(4 k^{3}

+ 12 k^{2}

+ 11 k

+ 3)

(k + 1)

⇥

Expression in progress	Divide leading terms	Quotient	Quotient × divisor	Remainder after subtracting
$4 k^{3}$ $+ 12 k^{2}$ $+ 11 k$ $+ 3$	$\frac{4 k^{3}}{k}$	$4 k^{2}$	$4 k^{2} \cdot (k + 1) = 4 k^{3} + 4 k^{2}$	$(4 k^{3}$ $+ 12 k^{2}$ $+ 11 k$ $+ 3) - (4 k^{3} + 4 k^{2}) = 8 k^{2} + 11 k + 3$
$8 k^{2} + 11 k + 3$	$\frac{8 k^{2}}{k}$	$8 k$	$8 k \cdot (k + 1) = 8 k^{2} + 8 k$	$(8 k^{2} + 11 k + 3) - (8 k^{2} + 8 k) = 3 k + 3$
$3 k + 3$	$\frac{3 k}{k}$	$3$	$3 \cdot (k + 1) = 3 k + 3$	$(3 k + 3) - (3 k + 3) = 0$

⇥So the result of the euclidean division is:

4 k^{2} + 8 k + 3

⇥1.3) Let's factorize in

ℝ

(4 k^{2} + 8 k + 3)

⇥Given the quadratic form

a k^{2} + b k + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 4

b = 8

and

c = 3

⇥

Δ = 8^{2} - 4 \cdot 4 \cdot 3

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

8^{2}

8 \cdot 8

64

■Multiplications are done before additions and subtractions:

4 \cdot 4

16

⇥

Δ = 64 - 16 \cdot 3

⇥ⓘ Multiplications are done before additions and subtractions:

16 \cdot 3

(10 + 6) \cdot 3

10 \cdot 3 + 6 \cdot 3

30 + 18

48

⇥

Δ = 64 - 48

⇥ⓘ Start from

64

and move back

48

on the number line:

64 - 48

16

⇥

Δ = 16

⇥As

Δ > 0

, the equation has two roots:

⇥The quadratic formula gives

k_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 8 - \sqrt{16}}{2 \cdot 4}

⇥

k_{1} = \frac{- 8 - \sqrt{16}}{2 \cdot 4}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

⇥ⅰ

2 \cdot 4

8

⇥

k_{1} = \frac{- 8 - 4}{8}

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

- 8 - 4

-12

⇥

k_{1} = - \frac{12}{8}

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

4

, their greatest common divisor:

- \frac{12}{8}

- \frac{3 \cdot 4}{2 \cdot 4}

- \frac{3}{2}

⇥

k_{1} = - \frac{3}{2}

⇥The quadratic formula gives

k_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 8 + \sqrt{16}}{2 \cdot 4}

⇥

k_{2} = \frac{- 8 + \sqrt{16}}{2 \cdot 4}

⇥ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

⇥ⅰ

2 \cdot 4

8

⇥

k_{2} = \frac{- 8 + 4}{8}

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

- 8 + 4

-4

⇥

k_{2} = - \frac{4}{8}

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

4

, their greatest common divisor:

- \frac{4}{8}

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

- \frac{1}{2}

⇥

k_{2} = - \frac{1}{2}

⇥A trinomial with roots

k_{1}

and

k_{2}

a \cdot (k - k_{1}) \cdot (k - k_{2})

; bringing the coefficient into the brackets:

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

(2 k + 1) \cdot (2 k + 3)

▷The factor

(4 k^{2} + 8 k + 3)

factors as:

(2 k + 1) \cdot (2 k + 3)

▶Factored expression:

(k + 1) \cdot (2 k + 1) \cdot (2 k + 3)