Factor the expression by grouping. First, the expression needs to be rewritten as $-2u^{2}+au+bu-28$. To find $a$ and $b$, set up a system to be solved.
$$a+b=15$$ $$ab=-2\left(-28\right)=56$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $56$.
$$1,56$$ $$2,28$$ $$4,14$$ $$7,8$$
Calculate the sum for each pair.
$$1+56=57$$ $$2+28=30$$ $$4+14=18$$ $$7+8=15$$
The solution is the pair that gives sum $15$.
$$a=8$$ $$b=7$$
Rewrite $-2u^{2}+15u-28$ as $\left(-2u^{2}+8u\right)+\left(7u-28\right)$.
$$\left(-2u^{2}+8u\right)+\left(7u-28\right)$$
Factor out $2u$ in the first and $-7$ in the second group.
$$2u\left(-u+4\right)-7\left(-u+4\right)$$
Factor out common term $-u+4$ by using distributive property.
$$\left(-u+4\right)\left(2u-7\right)$$
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 $ax^{2}+bx+c=0$.
$$-2u^{2}+15u-28=0$$
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 $u=\frac{-15±1}{-4}$ when $±$ is plus. Add $-15$ to $1$.
$$u=-\frac{14}{-4}$$
Reduce the fraction $\frac{-14}{-4}$ to lowest terms by extracting and canceling out $2$.
$$u=\frac{7}{2}$$
Now solve the equation $u=\frac{-15±1}{-4}$ when $±$ is minus. Subtract $1$ from $-15$.
$$u=-\frac{16}{-4}$$
Divide $-16$ by $-4$.
$$u=4$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{7}{2}$ for $x_{1}$ and $4$ for $x_{2}$.
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 $x^2+Bx+C=0$.
$$x ^ 2 -\frac{15}{2}x +14 = 0$$
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 $rs = C$
$$r + s = \frac{15}{2} $$ $$ rs = 14$$
Two numbers $r$ and $s$ sum up to $\frac{15}{2}$ exactly when the average of the two numbers is $\frac{1}{2}*\frac{15}{2} = \frac{15}{4}$. 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 $u$.
$$r = \frac{15}{4} - u$$ $$s = \frac{15}{4} + u$$
To solve for unknown quantity $u$, substitute these in the product equation $rs = 14$
$$(\frac{15}{4} - u) (\frac{15}{4} + u) = 14$$
Simplify by expanding $(a -b) (a + b) = a^2 – b^2$
$$\frac{225}{16} - u^2 = 14$$
Simplify the expression by subtracting $\frac{225}{16}$ on both sides
$$-u^2 = 14-\frac{225}{16} = -\frac{1}{16}$$
Simplify the expression by multiplying $-1$ on both sides and take the square root to obtain the value of unknown variable $u$
