Solve -3x^2+8x-5

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

Factor

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

Solution Steps

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

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

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

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

Factor out -x in -3x^{2}+5x.

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

Factor out common term 3x-5 by using distributive property.

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

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.

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

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.

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

Square 8.

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

Multiply -4 times -3.

$$x=\frac{-8±\sqrt{64+12\left(-5\right)}}{2\left(-3\right)}$$

Multiply 12 times -5.

$$x=\frac{-8±\sqrt{64-60}}{2\left(-3\right)}$$

Add 64 to -60.

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

Take the square root of 4.

$$x=\frac{-8±2}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-8±2}{-6}$$

Now solve the equation x=\frac{-8±2}{-6} when ± is plus. Add -8 to 2.

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

Divide -6 by -6.

$$x=1$$

Now solve the equation x=\frac{-8±2}{-6} when ± is minus. Subtract 2 from -8.

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

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

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

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

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

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

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

Cancel out 3, the greatest common factor in -3 and 3.

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

Evaluate

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