Math, asked by tharun2626, 1 year ago

1) 15x^2+4x-4< or = 0 solve by Algebraic method ?

Answers

Answered by Tasu06
1

Algebra Factoring Completely

1 Answer

Tony B

May 8, 2016

Answer:

(

3

x

+

2

)

(

5

x

2

)

Explanation:

It is a matter of spotting possible combinations and trying them out. They can be tried out on paper or in your head.

Factors of 15

{1,15} ; {3,5}

Factors of 4

{1,4} ; {2,2}

Cross link factors of 4 with factors of 15

'......................................................................

Consider: {1,4} linked to {1,15}

(

1

×

15

)

(

4

×

1

)

=

11

So this is no good for obtaining

4

x

'........................................................................

Consider: {2,2} linked to {1,15}

(

2

×

15

)

(

2

×

1

)

=

28

So this is no good for obtaining

4

x

'.......................................................................

Consider: {1,4} linked to {3,5}

(

3

×

4

)

(

1

×

5

)

=

7

So this is no good for obtaining

4

x

'..............................................................

Consider: {2,2} linked to {3,5}

(

2

×

5

)

(

2

×

3

)

=

4

this is the combination we need

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(

3

x

+

2

)

(

5

x

2

)

Answered by SagarKhawas
1

Explanation:

It is a matter of spotting possible combinations and trying them out. They can be tried out on paper or in your head.

Factors of 15

{1,15} ; {3,5}

Factors of 4

{1,4} ; {2,2}

Cross link factors of 4 with factors of 15

'......................................................................

Consider: {1,4} linked to {1,15}

(

1

×

15

)

(

4

×

1

)

=

11

So this is no good for obtaining

4

x

'........................................................................

Consider: {2,2} linked to {1,15}

(

2

×

15

)

(

2

×

1

)

=

28

So this is no good for obtaining

4

x

'.......................................................................

Consider: {1,4} linked to {3,5}

(

3

×

4

)

(

1

×

5

)

=

7

So this is no good for obtaining

4

x

'..............................................................

Consider: {2,2} linked to {3,5}

(

2

×

5

)

(

2

×

3

)

=

4

this is the combination we need

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(

3

x

+

2

)

(

5

x

2

)

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