Question

factorise x^3+6x^2+5-12
please answer soon
I will mark u as brainliest​

Answer 1

I have one example so, see that example and try to solve you will get ans..!

⇒

(

x

+

1

)

(

x

−

3

)

(

x

−

4

)

Factors are

x

=

−

1

,

3

,

4

Explanation:

(

x

3

−

6

x

2

+

5

x

+

12

)

Since the sum of the coefficients of the terms is zero,

x

+

1

is a factor.

Using synthetic division,

a

a

a

−

1

a

a

a

∣

a

a

a

1

a

a

a

−

6

a

a

a

5

a

a

a

12

a

a

a

a

a

a

a

a

a

∣

a

a

a

⏐

↓

a

a

−

1

a

a

a

7

a

−

12

a

a

a

a

a

a

a

a

a

∣

…

…

…

…

…

…

…

…

…

a

a

a

a

a

a

a

a

a

a

a

a

a

a

1

a

a

a

−

7

a

a

a

12

a

a

a

0

⇒

(

x

+

1

)

(

x

2

−

7

x

+

12

)

⇒

(

x

+

1

)

(

x

2

−

3

x

−

4

x

+

12

)

⇒

(

x

+

1

)

(

x

(

x

−

3

)

−

4

(

x

−

3

)

)

⇒

(

x

+

1

)

(

x

−

3

)

(

x

−

4

)

Factors are

x

=

−

1

,

3

,

4

Answer 2

Answer:

$\huge\underline{\overline{\mid {\bold {\green {☆ansWer☆}}\mid}}}$

${x}^{3} + {6x}^{2} + 5 - 12 = 0$

$= > {x}^{2} + {6x}^{2} + 5 = 12$

$= > {x}^{2} + {6x}^{2} = 12 - 5$

$= > {7x}^{2} = 7$

$= > {x}^{2} = \frac{7}{7}$

$= > {x}^{2} = 1$

$= > x = \sqrt{1}$