Solve -2u^2+11u+21

$$- 2 u ^ { 2 } + 11 u + 21$$

Factor

$$\left(7-u\right)\left(2u+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -2u^{2}+au+bu+21. To find a and b, set up a system to be solved.

$$a+b=11 ab=-2\times 21=-42$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -42.

$$-1,42 -2,21 -3,14 -6,7$$

Calculate the sum for each pair.

$$-1+42=41 -2+21=19 -3+14=11 -6+7=1$$

The solution is the pair that gives sum 11.

$$a=14 b=-3$$

Rewrite -2u^{2}+11u+21 as \left(-2u^{2}+14u\right)+\left(-3u+21\right).

$$\left(-2u^{2}+14u\right)+\left(-3u+21\right)$$

Factor out 2u in the first and 3 in the second group.

$$2u\left(-u+7\right)+3\left(-u+7\right)$$

Factor out common term -u+7 by using distributive property.

$$\left(-u+7\right)\left(2u+3\right)$$

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.

$$-2u^{2}+11u+21=0$$

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.

$$u=\frac{-11±\sqrt{11^{2}-4\left(-2\right)\times 21}}{2\left(-2\right)}$$

Square 11.

$$u=\frac{-11±\sqrt{121-4\left(-2\right)\times 21}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$u=\frac{-11±\sqrt{121+8\times 21}}{2\left(-2\right)}$$

Multiply 8 times 21.

$$u=\frac{-11±\sqrt{121+168}}{2\left(-2\right)}$$

Add 121 to 168.

$$u=\frac{-11±\sqrt{289}}{2\left(-2\right)}$$

Take the square root of 289.

$$u=\frac{-11±17}{2\left(-2\right)}$$

Multiply 2 times -2.

$$u=\frac{-11±17}{-4}$$

Now solve the equation u=\frac{-11±17}{-4} when ± is plus. Add -11 to 17.

$$u=\frac{6}{-4}$$

Reduce the fraction \frac{6}{-4} to lowest terms by extracting and canceling out 2.

$$u=-\frac{3}{2}$$

Now solve the equation u=\frac{-11±17}{-4} when ± is minus. Subtract 17 from -11.

$$u=-\frac{28}{-4}$$

Divide -28 by -4.

$$u=7$$

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 7 for x_{2}.

$$-2u^{2}+11u+21=-2\left(u-\left(-\frac{3}{2}\right)\right)\left(u-7\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$-2u^{2}+11u+21=-2\left(u+\frac{3}{2}\right)\left(u-7\right)$$

Add \frac{3}{2} to u by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-2u^{2}+11u+21=-2\times \frac{-2u-3}{-2}\left(u-7\right)$$

Cancel out 2, the greatest common factor in -2 and 2.

$$-2u^{2}+11u+21=\left(-2u-3\right)\left(u-7\right)$$

Steps Using Direct Factoring Method

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.

$$x ^ 2 -\frac{11}{2}x -\frac{21}{2} = 0$$

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

$$r + s = \frac{11}{2} rs = -\frac{21}{2}$$

Two numbers r and s sum up to \frac{11}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{2} = \frac{11}{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. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{11}{4} - u s = \frac{11}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{21}{2}

$$(\frac{11}{4} - u) (\frac{11}{4} + u) = -\frac{21}{2}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{121}{16} - u^2 = -\frac{21}{2}$$

Simplify the expression by subtracting \frac{121}{16} on both sides

$$-u^2 = -\frac{21}{2}-\frac{121}{16} = -\frac{289}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{289}{16} u = \pm\sqrt{\frac{289}{16}} = \pm \frac{17}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{11}{4} - \frac{17}{4} = -1.500 s = \frac{11}{4} + \frac{17}{4} = 7$$

Evaluate

$$\left(7-u\right)\left(2u+3\right)$$