Solve 10x+3-8x^2

$$10 x + 3 - 8 x ^ { 2 }$$

Factor

$$-\left(2x-3\right)\left(4x+1\right)$$

Solution Steps

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

$$-8x^{2}+10x+3$$

Factor the expression by grouping. First, the expression needs to be rewritten as -8x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=10 ab=-8\times 3=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 10.

$$a=12 b=-2$$

Rewrite -8x^{2}+10x+3 as \left(-8x^{2}+12x\right)+\left(-2x+3\right).

$$\left(-8x^{2}+12x\right)+\left(-2x+3\right)$$

Factor out -4x in the first and -1 in the second group.

$$-4x\left(2x-3\right)-\left(2x-3\right)$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-4x-1\right)$$

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.

$$-8x^{2}+10x+3=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-8\right)\times 3}}{2\left(-8\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-8\right)\times 3}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-10±\sqrt{100+32\times 3}}{2\left(-8\right)}$$

Multiply 32 times 3.

$$x=\frac{-10±\sqrt{100+96}}{2\left(-8\right)}$$

Add 100 to 96.

$$x=\frac{-10±\sqrt{196}}{2\left(-8\right)}$$

Take the square root of 196.

$$x=\frac{-10±14}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-10±14}{-16}$$

Now solve the equation x=\frac{-10±14}{-16} when ± is plus. Add -10 to 14.

$$x=\frac{4}{-16}$$

Reduce the fraction \frac{4}{-16} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-10±14}{-16} when ± is minus. Subtract 14 from -10.

$$x=-\frac{24}{-16}$$

Reduce the fraction \frac{-24}{-16} to lowest terms by extracting and canceling out 8.

$$x=\frac{3}{2}$$

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 \frac{3}{2} for x_{2}.

$$-8x^{2}+10x+3=-8\left(x-\left(-\frac{1}{4}\right)\right)\left(x-\frac{3}{2}\right)$$

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

$$-8x^{2}+10x+3=-8\left(x+\frac{1}{4}\right)\left(x-\frac{3}{2}\right)$$

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

$$-8x^{2}+10x+3=-8\times \frac{-4x-1}{-4}\left(x-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-8x^{2}+10x+3=-8\times \frac{-4x-1}{-4}\times \frac{-2x+3}{-2}$$

Multiply \frac{-4x-1}{-4} times \frac{-2x+3}{-2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-8x^{2}+10x+3=-8\times \frac{\left(-4x-1\right)\left(-2x+3\right)}{-4\left(-2\right)}$$

Multiply -4 times -2.

$$-8x^{2}+10x+3=-8\times \frac{\left(-4x-1\right)\left(-2x+3\right)}{8}$$

Cancel out 8, the greatest common factor in -8 and 8.

$$-8x^{2}+10x+3=-\left(-4x-1\right)\left(-2x+3\right)$$

Evaluate

$$3+10x-8x^{2}$$