Question

Factorisation of Quadartic polynomial. 2y^2+3y+1 . plz​

Answer 1

☆Answer☆

=》 2y² + 3y + 1

=》 2y² + 2y + y + 1

=》 2y ( y + 1 ) + 1 ( y + 1 )

=》 ( 2y + 1 )( y + 1 )

Answer 2

hey

answer..

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "y2" was replaced by "y^2".

Step by step solution :

STEP

1

:

Equation at the end of step 1

(2y2 - 3y) - 1 = 0

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring 2y2-3y-1

The first term is, 2y2 its coefficient is 2 .

The middle term is, -3y its coefficient is -3 .

The last term, "the constant", is -1

Step-1 : Multiply the coefficient of the first term by the constant 2 • -1 = -2

Step-2 : Find two factors of -2 whose sum equals the coefficient of the middle term, which is -3 .

-2 + 1 = -1

-1 + 2 = 1

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step

2

:

2y2 - 3y - 1 = 0

STEP

3

:

Parabola, Finding the Vertex:

3.1 Find the Vertex of t = 2y2-3y-1

For any parabola,Ay2+By+C,the y -coordinate of the vertex is given by -B/(2A) . In our case the y coordinate is 0.7500

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