How do you factor #-2x ^ 2 + 7x -3?

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Answer 1 · 2015-05-07T05:03:54

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
$N_{1} + N_{2} = b = 7$ $N_1 +N_2 = b = 7$

After trying out a few numbers we get $N_{1} = 6$ $N_1 = 6$ and $N_{2} = 1$ $N_2 =1$

$6 \cdot 1 = 6$ $6*1 = 6$ , and $6 + 1 = 7$ $6+1= 7$

$- 2 x^{2} + 7 x - 3 = - 2 x^{2} + 6 x + x - 3$ $-2x ^ 2 + 7x -3 = -2x^2 +6x + x - 3$

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

$x - 3$ $x-3$ is a common factor to each of the terms

$= (x - 3) (- 2 x + 1)$ $=color(green)((x-3)(-2x+1)$