Solve 14x^2+35x

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

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.

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

Take the square root of 35^{2}.

$$x=\frac{-35±35}{2\times 14}$$

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.

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

Divide 0 by 28.

$$x=0$$

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

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

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

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

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

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

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

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

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

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

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

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

Evaluate

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