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Question # Decide equation using only factors to solve this problem 12x^{2} + 5x - 2 = 0

Comparing two groups
ANSWERED Decide equation using only factors to solve this problem
$$12x^{2}\ +\ 5x\ -\ 2 = 0$$ 2021-03-10

In order to factor the quadratic, we need to find the values of a, b and c for the given quadratic $$12x^{2}\ +\ 5x\ -\ 2 = 0$$ on comparing with standard quadratic
$$ax^{2}\ +\ bx\ +\ c = 0$$
$$12x^{2}\ +\ 5x\ -\ 2 = 0$$
$$a = 12\, b = 5,\ c =\ -2.$$
Now, let us apply ac method to factor the quadratic.
We got $$ac=\ -24\ and\ b=5.$$ So, we need to find the factors of -24, those add upto 5.
$$ac = 12 \cdot\ -2 =\ -24$$(Product)
$$b = 5$$(Sum)
-24 could be factored in two numbers 8 and -3 that add upto 5.
Therefore, middle term 5x of expression $$12x^{2}\ +\ 5x\ -\ 2 =0,$$ could be break into two terms 8x and -3x.
Then we need to make the break the expression into groups.
Factor out 3x form $$12x^{2}\ -\ 3x$$ and factor out 2 from:
$$8x\ -\ 2.$$
Then factor out common parenthesis $$(4x\ -\ 1).$$
We got factored form $$(4x\ -\ 1) (3x\ +\ 2) = 0$$
$$12x^{2}\ +\ 5x\ -\ 2 = 0$$
$$(12x^{2}\ -\ 3x)\ +\ (8x\ -\ 2) = 0$$
$$3x(4x\ -\ 1)\ +\ 2 (4x\ -\ 1) = 0$$
Factor out common parenthesis $$(4x\ -\ 1)$$
$$(4x\ -\ 1)(3x\ +\ 2) = 0$$
By applying zero product rule, we need to set each factor equal to zero and solve for x.
We got $$x = \frac{1}{4},\ \frac{-2}{3}.$$
$$4x\ -\ 1 = 0\ or\ 3x\ +\ 2 = 0$$
$$+1 +1 -2 -2$$
$$\frac{4x}{4} = \frac{1}{4}\ or\ \frac{3x}{3} = \frac{-2}{3}$$
$$x = 1.4\ or\ x = \frac{-2}{3}$$