Factor the expression by grouping. First, the expression needs to be rewritten as $4x^{2}+ax+bx+3$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-8$$ $$ab=4\times 3=12$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $12$.
$$-1,-12$$ $$-2,-6$$ $$-3,-4$$
Calculate the sum for each pair.
$$-1-12=-13$$ $$-2-6=-8$$ $$-3-4=-7$$
The solution is the pair that gives sum $-8$.
$$a=-6$$ $$b=-2$$
Rewrite $4x^{2}-8x+3$ as $\left(4x^{2}-6x\right)+\left(-2x+3\right)$.
$$\left(4x^{2}-6x\right)+\left(-2x+3\right)$$
Factor out $2x$ in the first and $-1$ in the second group.
$$2x\left(2x-3\right)-\left(2x-3\right)$$
Factor out common term $2x-3$ by using distributive property.
$$\left(2x-3\right)\left(2x-1\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$4x^{2}-8x+3=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
Now solve the equation $x=\frac{8±4}{8}$ when $±$ is plus. Add $8$ to $4$.
$$x=\frac{12}{8}$$
Reduce the fraction $\frac{12}{8}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{3}{2}$$
Now solve the equation $x=\frac{8±4}{8}$ when $±$ is minus. Subtract $4$ from $8$.
$$x=\frac{4}{8}$$
Reduce the fraction $\frac{4}{8}$ to lowest terms by extracting and canceling out $4$.
$$x=\frac{1}{2}$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{3}{2}$ for $x_{1}$ and $\frac{1}{2}$ for $x_{2}$.
Multiply $\frac{2x-3}{2}$ times $\frac{2x-1}{2}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form $x^2+Bx+C=0$.This is achieved by dividing both sides of the equation by $4$
$$x ^ 2 -2x +\frac{3}{4} = 0$$
Let $r$ and $s$ be the factors for the quadratic equation such that $x^2+Bx+C=(x−r)(x−s)$ where sum of factors $(r+s)=−B$ and the product of factors $rs = C$
$$r + s = 2 $$ $$ rs = \frac{3}{4}$$
Two numbers $r$ and $s$ sum up to $2$ exactly when the average of the two numbers is $\frac{1}{2}*2 = 1$. You can also see that the midpoint of $r$ and $s$ corresponds to the axis of symmetry of the parabola represented by the quadratic equation $y=x^2+Bx+C$. The values of $r$ and $s$ are equidistant from the center by an unknown quantity $u$. Express $r$ and $s$ with respect to variable $u$.
$$r = 1 - u$$ $$s = 1 + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = \frac{3}{4}$
$$(1 - u) (1 + u) = \frac{3}{4}$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$1 - u^2 = \frac{3}{4}$$
Simplify the expression by subtracting $1$ on both sides
$$-u^2 = \frac{3}{4}-1 = -\frac{1}{4}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
