Solve -5x^2+8x-3

$$-5 { x }^{ 2 } +8x-3$$

Factor

$$\left(1-x\right)\left(5x-3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as -5x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=8 ab=-5\left(-3\right)=15$$

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 15.

$$1,15 3,5$$

Calculate the sum for each pair.

$$1+15=16 3+5=8$$

The solution is the pair that gives sum 8.

$$a=5 b=3$$

Rewrite -5x^{2}+8x-3 as \left(-5x^{2}+5x\right)+\left(3x-3\right).

$$\left(-5x^{2}+5x\right)+\left(3x-3\right)$$

Factor out 5x in the first and -3 in the second group.

$$5x\left(-x+1\right)-3\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(5x-3\right)$$

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.

$$-5x^{2}+8x-3=0$$

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.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-8±\sqrt{64+20\left(-3\right)}}{2\left(-5\right)}$$

Multiply 20 times -3.

$$x=\frac{-8±\sqrt{64-60}}{2\left(-5\right)}$$

Add 64 to -60.

$$x=\frac{-8±\sqrt{4}}{2\left(-5\right)}$$

Take the square root of 4.

$$x=\frac{-8±2}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-8±2}{-10}$$

Now solve the equation x=\frac{-8±2}{-10} when ± is plus. Add -8 to 2.

$$x=-\frac{6}{-10}$$

Reduce the fraction \frac{-6}{-10} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{5}$$

Now solve the equation x=\frac{-8±2}{-10} when ± is minus. Subtract 2 from -8.

$$x=-\frac{10}{-10}$$

Divide -10 by -10.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{5} for x_{1} and 1 for x_{2}.

$$-5x^{2}+8x-3=-5\left(x-\frac{3}{5}\right)\left(x-1\right)$$

Subtract \frac{3}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-5x^{2}+8x-3=-5\times \frac{-5x+3}{-5}\left(x-1\right)$$

Cancel out 5, the greatest common factor in -5 and 5.

$$-5x^{2}+8x-3=\left(-5x+3\right)\left(x-1\right)$$

Evaluate

$$\left(1-x\right)\left(5x-3\right)$$