Solve 3+17x-28x^2)

$$3 + 17 x - 28 x ^ { 2 } )$$

Factor

$$-\left(4x-3\right)\left(7x+1\right)$$

Solution Steps

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

$$-28x^{2}+17x+3$$

Factor the expression by grouping. First, the expression needs to be rewritten as -28x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=17 ab=-28\times 3=-84$$

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 -84.

$$-1,84 -2,42 -3,28 -4,21 -6,14 -7,12$$

Calculate the sum for each pair.

$$-1+84=83 -2+42=40 -3+28=25 -4+21=17 -6+14=8 -7+12=5$$

The solution is the pair that gives sum 17.

$$a=21 b=-4$$

Rewrite -28x^{2}+17x+3 as \left(-28x^{2}+21x\right)+\left(-4x+3\right).

$$\left(-28x^{2}+21x\right)+\left(-4x+3\right)$$

Factor out -7x in the first and -1 in the second group.

$$-7x\left(4x-3\right)-\left(4x-3\right)$$

Factor out common term 4x-3 by using distributive property.

$$\left(4x-3\right)\left(-7x-1\right)$$

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.

$$-28x^{2}+17x+3=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\left(-28\right)\times 3}}{2\left(-28\right)}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\left(-28\right)\times 3}}{2\left(-28\right)}$$

Multiply -4 times -28.

$$x=\frac{-17±\sqrt{289+112\times 3}}{2\left(-28\right)}$$

Multiply 112 times 3.

$$x=\frac{-17±\sqrt{289+336}}{2\left(-28\right)}$$

Add 289 to 336.

$$x=\frac{-17±\sqrt{625}}{2\left(-28\right)}$$

Take the square root of 625.

$$x=\frac{-17±25}{2\left(-28\right)}$$

Multiply 2 times -28.

$$x=\frac{-17±25}{-56}$$

Now solve the equation x=\frac{-17±25}{-56} when ± is plus. Add -17 to 25.

$$x=\frac{8}{-56}$$

Reduce the fraction \frac{8}{-56} to lowest terms by extracting and canceling out 8.

$$x=-\frac{1}{7}$$

Now solve the equation x=\frac{-17±25}{-56} when ± is minus. Subtract 25 from -17.

$$x=-\frac{42}{-56}$$

Reduce the fraction \frac{-42}{-56} to lowest terms by extracting and canceling out 14.

$$x=\frac{3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{7} for x_{1} and \frac{3}{4} for x_{2}.

$$-28x^{2}+17x+3=-28\left(x-\left(-\frac{1}{7}\right)\right)\left(x-\frac{3}{4}\right)$$

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

$$-28x^{2}+17x+3=-28\left(x+\frac{1}{7}\right)\left(x-\frac{3}{4}\right)$$

Add \frac{1}{7} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$-28x^{2}+17x+3=-28\times \frac{-7x-1}{-7}\left(x-\frac{3}{4}\right)$$

Subtract \frac{3}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-28x^{2}+17x+3=-28\times \frac{-7x-1}{-7}\times \frac{-4x+3}{-4}$$

Multiply \frac{-7x-1}{-7} times \frac{-4x+3}{-4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-28x^{2}+17x+3=-28\times \frac{\left(-7x-1\right)\left(-4x+3\right)}{-7\left(-4\right)}$$

Multiply -7 times -4.

$$-28x^{2}+17x+3=-28\times \frac{\left(-7x-1\right)\left(-4x+3\right)}{28}$$

Cancel out 28, the greatest common factor in -28 and 28.

$$-28x^{2}+17x+3=-\left(-7x-1\right)\left(-4x+3\right)$$

Evaluate

$$3+17x-28x^{2}$$