Solve -4x^2+11x+3

$$- 4 x ^ { 2 } + 11 x + 3$$

Factor

$$\left(3-x\right)\left(4x+1\right)$$

Solution Steps

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.

$$a+b=11 ab=-4\times 3=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 11.

$$a=12 b=-1$$

Rewrite -4x^{2}+11x+3 as \left(-4x^{2}+12x\right)+\left(-x+3\right).

$$\left(-4x^{2}+12x\right)+\left(-x+3\right)$$

Factor out 4x in -4x^{2}+12x.

$$4x\left(-x+3\right)-x+3$$

Factor out common term -x+3 by using distributive property.

$$\left(-x+3\right)\left(4x+1\right)$$

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.

$$-4x^{2}+11x+3=0$$

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.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-4\right)\times 3}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-11±\sqrt{121+16\times 3}}{2\left(-4\right)}$$

Multiply 16 times 3.

$$x=\frac{-11±\sqrt{121+48}}{2\left(-4\right)}$$

Add 121 to 48.

$$x=\frac{-11±\sqrt{169}}{2\left(-4\right)}$$

Take the square root of 169.

$$x=\frac{-11±13}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-11±13}{-8}$$

Now solve the equation x=\frac{-11±13}{-8} when ± is plus. Add -11 to 13.

$$x=\frac{2}{-8}$$

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

$$x=-\frac{1}{4}$$

Now solve the equation x=\frac{-11±13}{-8} when ± is minus. Subtract 13 from -11.

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

Divide -24 by -8.

$$x=3$$

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

$$-4x^{2}+11x+3=-4\left(x-\left(-\frac{1}{4}\right)\right)\left(x-3\right)$$

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

$$-4x^{2}+11x+3=-4\left(x+\frac{1}{4}\right)\left(x-3\right)$$

Add \frac{1}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+11x+3=-4\times \frac{-4x-1}{-4}\left(x-3\right)$$

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

$$-4x^{2}+11x+3=\left(-4x-1\right)\left(x-3\right)$$

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.

$$x ^ 2 -\frac{11}{4}x -\frac{3}{4} = 0$$

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

$$r + s = \frac{11}{4} rs = -\frac{3}{4}$$

Two numbers r and s sum up to \frac{11}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{4} = \frac{11}{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>

$$r = \frac{11}{8} - u s = \frac{11}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}

$$(\frac{11}{8} - u) (\frac{11}{8} + u) = -\frac{3}{4}$$

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

$$\frac{121}{64} - u^2 = -\frac{3}{4}$$

Simplify the expression by subtracting \frac{121}{64} on both sides

$$-u^2 = -\frac{3}{4}-\frac{121}{64} = -\frac{169}{64}$$

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

$$u^2 = \frac{169}{64} u = \pm\sqrt{\frac{169}{64}} = \pm \frac{13}{8}$$

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

$$r =\frac{11}{8} - \frac{13}{8} = -0.250 s = \frac{11}{8} + \frac{13}{8} = 3$$

Evaluate

$$\left(3-x\right)\left(4x+1\right)$$