Best Answer

There is no rational factorisation of 6y2 + 19y - 13

The roots are [-19 +/- sqrt(673)]/12

which are -3.74519 and 0.57852

Q: Factor the trinomial 6y2 plus 19y-13?

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3y^2(5y + 2)

2(3y + 1)(y - 2)

A = 3x2 + 9xy + 6y2l = 3x + 6yA = l×w∴ w = A/l∴ w = (3x2 + 9xy + 6y2) / (3x + 6y)∴ w = x + 1

It is a quadratic expression in the variable x.

3x2 - 2y2 = 9x2 + 4y2 - 12xy (subtract 3x2 and add 2y2 to both sides) 0 = 9x2 -3x2 + 4y2 + 2y2 - 12xy 0 = 6x2 + 6y2 - 12xy (divide by 6 to both sides) 0 = x2 + y2 - 12xy or x2 - 2xy + y2 = 0 (x - y)(x - y) = 0 x - y = 0 x = y

Related questions

3(2y + 3)(y + 2)

20x2 + 22xy + 6y2 = 20x2 + 10xy + 12xy + 6y2 = 10x(2x + y) + 6y(2x + y) = (2x + y)(10x + 6y) = 2(2x + y)(5x + 3y)

6y2+17y-14 = (2x+7)(3x-2) when factored

It is: 3y and y

It is: (6y+1)(y-5) when factored

2(3y + 1)(y - 2)

3y^2(5y + 2)

A = 3x2 + 9xy + 6y2l = 3x + 6yA = l×w∴ w = A/l∴ w = (3x2 + 9xy + 6y2) / (3x + 6y)∴ w = x + 1

6y2 + 11y - 2 = (6y - 1) (y + 2)

6y2 - 36 = 6ySubtract 6y from each side:6y2 - 6y - 36 = 0Divide each side by 6:y2 - y - 6 = 0Factor the expression on the left side:(y + 2) (y - 3) = 0The equation is a true statement if either factor is = 0.y + 2 = 0 ==> y = -2y - 3 = 0 ==> y = 3

6y2

(3x+2y)*(5x-3y) =3x*5x + 3x*(-3y) + 2y*5x + 2y*(-3y) [This is sometimes called the FOIL method - You multiply the First members of each bracket, then the Outers, Inners, Lasts] = 15x2 - 9xy + 10xy - 6y2 = 15x2 + xy - 6y2