Solve 16x-7-4x^2

$$16x-7-4 { x }^{ 2 }$$

Factor

$$\left(1-2x\right)\left(2x-7\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-4x^{2}+16x-7$$

Factor the expression by grouping. First, the expression needs to be rewritten as -4x^{2}+ax+bx-7. To find a and b, set up a system to be solved.

$$a+b=16 ab=-4\left(-7\right)=28$$

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 28.

$$1,28 2,14 4,7$$

Calculate the sum for each pair.

$$1+28=29 2+14=16 4+7=11$$

The solution is the pair that gives sum 16.

$$a=14 b=2$$

Rewrite -4x^{2}+16x-7 as \left(-4x^{2}+14x\right)+\left(2x-7\right).

$$\left(-4x^{2}+14x\right)+\left(2x-7\right)$$

Factor out -2x in -4x^{2}+14x.

$$-2x\left(2x-7\right)+2x-7$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(-2x+1\right)$$

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.

$$-4x^{2}+16x-7=0$$

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.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-4\right)\left(-7\right)}}{2\left(-4\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-4\right)\left(-7\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-16±\sqrt{256+16\left(-7\right)}}{2\left(-4\right)}$$

Multiply 16 times -7.

$$x=\frac{-16±\sqrt{256-112}}{2\left(-4\right)}$$

Add 256 to -112.

$$x=\frac{-16±\sqrt{144}}{2\left(-4\right)}$$

Take the square root of 144.

$$x=\frac{-16±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-16±12}{-8}$$

Now solve the equation x=\frac{-16±12}{-8} when ± is plus. Add -16 to 12.

$$x=-\frac{4}{-8}$$

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-16±12}{-8} when ± is minus. Subtract 12 from -16.

$$x=-\frac{28}{-8}$$

Reduce the fraction \frac{-28}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{7}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{2} for x_{1} and \frac{7}{2} for x_{2}.

$$-4x^{2}+16x-7=-4\left(x-\frac{1}{2}\right)\left(x-\frac{7}{2}\right)$$

Subtract \frac{1}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+16x-7=-4\times \frac{-2x+1}{-2}\left(x-\frac{7}{2}\right)$$

Subtract \frac{7}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+16x-7=-4\times \frac{-2x+1}{-2}\times \frac{-2x+7}{-2}$$

Multiply \frac{-2x+1}{-2} times \frac{-2x+7}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+16x-7=-4\times \frac{\left(-2x+1\right)\left(-2x+7\right)}{-2\left(-2\right)}$$

Multiply -2 times -2.

$$-4x^{2}+16x-7=-4\times \frac{\left(-2x+1\right)\left(-2x+7\right)}{4}$$

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

$$-4x^{2}+16x-7=-\left(-2x+1\right)\left(-2x+7\right)$$

Evaluate

$$\left(1-2x\right)\left(2x-7\right)$$