Solve 4n^2+n-33

$$4 n ^ { 2 } + n - 33$$

Factor

$$\left(4n-11\right)\left(n+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4n^{2}+an+bn-33. To find a and b, set up a system to be solved.

$$a+b=1 ab=4\left(-33\right)=-132$$

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 -132.

$$-1,132 -2,66 -3,44 -4,33 -6,22 -11,12$$

Calculate the sum for each pair.

$$-1+132=131 -2+66=64 -3+44=41 -4+33=29 -6+22=16 -11+12=1$$

The solution is the pair that gives sum 1.

$$a=-11 b=12$$

Rewrite 4n^{2}+n-33 as \left(4n^{2}-11n\right)+\left(12n-33\right).

$$\left(4n^{2}-11n\right)+\left(12n-33\right)$$

Factor out n in the first and 3 in the second group.

$$n\left(4n-11\right)+3\left(4n-11\right)$$

Factor out common term 4n-11 by using distributive property.

$$\left(4n-11\right)\left(n+3\right)$$

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.

$$4n^{2}+n-33=0$$

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.

$$n=\frac{-1±\sqrt{1^{2}-4\times 4\left(-33\right)}}{2\times 4}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\times 4\left(-33\right)}}{2\times 4}$$

Multiply -4 times 4.

$$n=\frac{-1±\sqrt{1-16\left(-33\right)}}{2\times 4}$$

Multiply -16 times -33.

$$n=\frac{-1±\sqrt{1+528}}{2\times 4}$$

Add 1 to 528.

$$n=\frac{-1±\sqrt{529}}{2\times 4}$$

Take the square root of 529.

$$n=\frac{-1±23}{2\times 4}$$

Multiply 2 times 4.

$$n=\frac{-1±23}{8}$$

Now solve the equation n=\frac{-1±23}{8} when ± is plus. Add -1 to 23.

$$n=\frac{22}{8}$$

Reduce the fraction \frac{22}{8} to lowest terms by extracting and canceling out 2.

$$n=\frac{11}{4}$$

Now solve the equation n=\frac{-1±23}{8} when ± is minus. Subtract 23 from -1.

$$n=-\frac{24}{8}$$

Divide -24 by 8.

$$n=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{11}{4} for x_{1} and -3 for x_{2}.

$$4n^{2}+n-33=4\left(n-\frac{11}{4}\right)\left(n-\left(-3\right)\right)$$

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

$$4n^{2}+n-33=4\left(n-\frac{11}{4}\right)\left(n+3\right)$$

Subtract \frac{11}{4} from n by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4n^{2}+n-33=4\times \frac{4n-11}{4}\left(n+3\right)$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4n^{2}+n-33=\left(4n-11\right)\left(n+3\right)$$

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.This is achieved by dividing both sides of the equation by 4

$$x ^ 2 +\frac{1}{4}x -\frac{33}{4} = 0$$

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

$$r + s = -\frac{1}{4} rs = -\frac{33}{4}$$

Two numbers r and s sum up to -\frac{1}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{4} = -\frac{1}{8}. 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>

$$r = -\frac{1}{8} - u s = -\frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{33}{4}

$$(-\frac{1}{8} - u) (-\frac{1}{8} + u) = -\frac{33}{4}$$

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

$$\frac{1}{64} - u^2 = -\frac{33}{4}$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = -\frac{33}{4}-\frac{1}{64} = -\frac{529}{64}$$

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

$$u^2 = \frac{529}{64} u = \pm\sqrt{\frac{529}{64}} = \pm \frac{23}{8}$$

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

$$r =-\frac{1}{8} - \frac{23}{8} = -3 s = -\frac{1}{8} + \frac{23}{8} = 2.750$$

Evaluate

$$\left(4n-11\right)\left(n+3\right)$$