How do you factor $2 x^{2} - 7 x + 3$ $2x^2 -7x +3$ ?

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Answer 1 · 2015-11-08T18:51:57

$2 x^{2} - 7 x + 3 = (2 x - 1) (x - 3)$ $2x^2-7x+3=(2x-1)(x-3)$

$2 x^{2} - 7 x + 3$ $2x^2-7x+3$ is a quadratic equation, $a x^{2} + b x + c$ $ax^2+bx+c$ , where $a = 2, b = - 7, and c = 3$ $a=2, b=-7, and c=3$ .

Use the $a \cdot c$ $a*c$ method to factor this equation (also called splitting the middle).

Multiply $a$ $a$ times $c$ $c$ .

$2 \times 3 = 6$ $2xx3=6$

Determine two numbers that when added equal $- 7$ $-7$ and when multiplied equal $6$ $6$ . The numbers $- x$ $-x$ and $- 6$ $-6$ fit the criteria.

Rewrite the equation substituting $- x$ $-x$ and $- 6 x$ $-6x$ for $- 7 x$ $-7x$ .

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

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

Factor out the common term $(2 x - 1)$ $(2x-1)$

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

How do you factor 2x^2 -7x +32x2−7x+32x^2 -7x +3?