How do you factor $2 x^{2} - 5 x + 3$ $2x^2-5x+3$ ?

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How do you factor $2 x^{2} - 5 x + 3$ $2x^2-5x+3$ ?

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Answer 1 · 2015-08-13T15:41:23

$(2 x - 3) (x - 1)$ $color(blue)((2x-3)(x-1)$ is the factorised form of the expression.

Explanation:

$2 x^{2} - 5 x + 3$ $2x^2 -5x +3$

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

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

$N_{1} \cdot N_{2} = a \cdot c = 2 \cdot 3 = 6$ $N_1*N_2 = a*c = 2*3 = 6$
and
$N_{1} + N_{2} = b = - 5$ $N_1 +N_2 = b = -5$

After trying out a few numbers we get $N_{1} = - 2$ $N_1 = -2$ and $N_{2} = - 3$ $N_2 =-3$
$(- 2) \cdot (- 3) = 6$ $(-2)*(-3) = 6$ , and $(- 2) + (- 3) = - 5$ $(-2)+(-3)= -5$

$2 x^{2} - 5 x + 3 = 2 x^{2} - 2 x - 3 x + 3$ $2x^2color(blue)( -5x) +3 = 2x^2 color(blue)(-2x-3x) +3$

$= 2 x (x - 1) - 3 (x - 1)$ $=2x(x-1) -3(x-1)$

$(2 x - 3) (x - 1)$ $color(blue)((2x-3)(x-1)$ is the factorised form of the expression.

Answer 2 · 2015-08-14T04:37:45

factor: $y = 2 x^{2} - 5 x + 3$ $y = 2x^2 - 5x + 3$

Ans: (x - 1)(2x - 3)

Explanation:

Since a + b + c = 0, use the shortcut. One factor is (x - 1) and the other is $(x - \frac{c}{a}) = (x - \frac{3}{2})$ $(x - c/a) = (x - 3/2)$
Factoring form: y = (x - 1)( 2x - 3)
The shortcut avoids the lengthy factoring by grouping.

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How do you factor $2 x^{2} - 5 x + 3$ $2x^2-5x+3$ ?

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How do you factor $2 x^{2} - 5 x + 3$ $2x^2-5x+3$ ?