How do you factor $3 y^{2} - 21 y + 36$ $3y^2 -21y+36$ ?

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Answer 1 · 2015-08-11T16:56:05

$(3 y - 9) (y - 4)$ $color(green)((3y-9)(y-4)$ is the factorised form of the expression.

$3 y^{2} - 21 y + 36$ $3y^2−21y+36$

We can Split the Middle Term of this expression to factorise it.

In this technique, if we have to factorise an expression like $a y^{2} + b y + c$ $ay^2 + by + c$ , we need to think of two numbers such that:

$N_{1} \cdot N_{2} = a \cdot c = 3 \cdot 36 = 108$ $N_1*N_2 = a*c = 3*36 = 108$

and,

$N_{1} + N_{2} = b = - 21$ $N_1 +N_2 = b = -21$

After trying out a few numbers we get $N_{1} = - 12$ $N_1 = -12$ and $N_{2} = - 9$ $N_2 =-9$

$(- 12) \cdot (- 9) = 108$ $(-12)*(-9) = 108$ , and $(- 12) + (- 9) = - 21$ $(-12)+(-9)= -21$

$3 y^{2} - 21 y + 36 = 3 y^{2} - 12 y - 9 y + 36$ $3y^2−color(green)(21y)+36 = 3y^2color(green)(-12y-9y)+36$

$= 3 y (y - 4) - 9 (y - 4)$ $=3y(y-4) -9(y-4)$

$(3 y - 9) (y - 4)$ $color(green)((3y-9)(y-4)$ is the factorised form of the expression.

How do you factor 3y2−21y+363y^2 -21y+36?