Solve 14x^2-35x

$$14 { x }^{ 2 } -35x$$

Factor

$$7x\left(2x-5\right)$$

Solution Steps

Factor out 7.

$$7\left(2x^{2}-5x\right)$$

Consider 2x^{2}-5x. Factor out x.

$$x\left(2x-5\right)$$

Rewrite the complete factored expression.

$$7x\left(2x-5\right)$$

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.

$$14x^{2}-35x=0$$

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.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}}}{2\times 14}$$

Take the square root of \left(-35\right)^{2}.

$$x=\frac{-\left(-35\right)±35}{2\times 14}$$

The opposite of -35 is 35.

$$x=\frac{35±35}{2\times 14}$$

Multiply 2 times 14.

$$x=\frac{35±35}{28}$$

Now solve the equation x=\frac{35±35}{28} when ± is plus. Add 35 to 35.

$$x=\frac{70}{28}$$

Reduce the fraction \frac{70}{28} to lowest terms by extracting and canceling out 14.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{35±35}{28} when ± is minus. Subtract 35 from 35.

$$x=\frac{0}{28}$$

Divide 0 by 28.

$$x=0$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and 0 for x_{2}.

$$14x^{2}-35x=14\left(x-\frac{5}{2}\right)x$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$14x^{2}-35x=14\times \frac{2x-5}{2}x$$

Cancel out 2, the greatest common factor in 14 and 2.

$$14x^{2}-35x=7\left(2x-5\right)x$$

Evaluate

$$7x\left(2x-5\right)$$