Solve 5x^2-16x-21

$$5 { x }^{ 2 } -16x-21$$

Factor

$$\left(5x-21\right)\left(x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=-16 ab=5\left(-21\right)=-105$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -105.

$$1,-105 3,-35 5,-21 7,-15$$

Calculate the sum for each pair.

$$1-105=-104 3-35=-32 5-21=-16 7-15=-8$$

The solution is the pair that gives sum -16.

$$a=-21 b=5$$

Rewrite 5x^{2}-16x-21 as \left(5x^{2}-21x\right)+\left(5x-21\right).

$$\left(5x^{2}-21x\right)+\left(5x-21\right)$$

Factor out x in 5x^{2}-21x.

$$x\left(5x-21\right)+5x-21$$

Factor out common term 5x-21 by using distributive property.

$$\left(5x-21\right)\left(x+1\right)$$

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.

$$5x^{2}-16x-21=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\left(-21\right)}}{2\times 5}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 5\left(-21\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-16\right)±\sqrt{256-20\left(-21\right)}}{2\times 5}$$

Multiply -20 times -21.

$$x=\frac{-\left(-16\right)±\sqrt{256+420}}{2\times 5}$$

Add 256 to 420.

$$x=\frac{-\left(-16\right)±\sqrt{676}}{2\times 5}$$

Take the square root of 676.

$$x=\frac{-\left(-16\right)±26}{2\times 5}$$

The opposite of -16 is 16.

$$x=\frac{16±26}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{16±26}{10}$$

Now solve the equation x=\frac{16±26}{10} when ± is plus. Add 16 to 26.

$$x=\frac{42}{10}$$

Reduce the fraction \frac{42}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{21}{5}$$

Now solve the equation x=\frac{16±26}{10} when ± is minus. Subtract 26 from 16.

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

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{21}{5} for x_{1} and -1 for x_{2}.

$$5x^{2}-16x-21=5\left(x-\frac{21}{5}\right)\left(x-\left(-1\right)\right)$$

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

$$5x^{2}-16x-21=5\left(x-\frac{21}{5}\right)\left(x+1\right)$$

Subtract \frac{21}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$5x^{2}-16x-21=5\times \frac{5x-21}{5}\left(x+1\right)$$

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

$$5x^{2}-16x-21=\left(5x-21\right)\left(x+1\right)$$

Evaluate

$$\left(5x-21\right)\left(x+1\right)$$