Solve 4a^2-19a-5

$$4 { a }^{ 2 } -19a-5$$

Factor

$$\left(a-5\right)\left(4a+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4a^{2}+pa+qa-5. To find p and q, set up a system to be solved.

$$p+q=-19 pq=4\left(-5\right)=-20$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -19.

$$p=-20 q=1$$

Rewrite 4a^{2}-19a-5 as \left(4a^{2}-20a\right)+\left(a-5\right).

$$\left(4a^{2}-20a\right)+\left(a-5\right)$$

Factor out 4a in 4a^{2}-20a.

$$4a\left(a-5\right)+a-5$$

Factor out common term a-5 by using distributive property.

$$\left(a-5\right)\left(4a+1\right)$$

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.

$$4a^{2}-19a-5=0$$

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.

$$a=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Square -19.

$$a=\frac{-\left(-19\right)±\sqrt{361-4\times 4\left(-5\right)}}{2\times 4}$$

Multiply -4 times 4.

$$a=\frac{-\left(-19\right)±\sqrt{361-16\left(-5\right)}}{2\times 4}$$

Multiply -16 times -5.

$$a=\frac{-\left(-19\right)±\sqrt{361+80}}{2\times 4}$$

Add 361 to 80.

$$a=\frac{-\left(-19\right)±\sqrt{441}}{2\times 4}$$

Take the square root of 441.

$$a=\frac{-\left(-19\right)±21}{2\times 4}$$

The opposite of -19 is 19.

$$a=\frac{19±21}{2\times 4}$$

Multiply 2 times 4.

$$a=\frac{19±21}{8}$$

Now solve the equation a=\frac{19±21}{8} when ± is plus. Add 19 to 21.

$$a=\frac{40}{8}$$

Divide 40 by 8.

$$a=5$$

Now solve the equation a=\frac{19±21}{8} when ± is minus. Subtract 21 from 19.

$$a=-\frac{2}{8}$$

Reduce the fraction \frac{-2}{8} to lowest terms by extracting and canceling out 2.

$$a=-\frac{1}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 5 for x_{1} and -\frac{1}{4} for x_{2}.

$$4a^{2}-19a-5=4\left(a-5\right)\left(a-\left(-\frac{1}{4}\right)\right)$$

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

$$4a^{2}-19a-5=4\left(a-5\right)\left(a+\frac{1}{4}\right)$$

Add \frac{1}{4} to a by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4a^{2}-19a-5=4\left(a-5\right)\times \frac{4a+1}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4a^{2}-19a-5=\left(a-5\right)\left(4a+1\right)$$

Evaluate

$$\left(a-5\right)\left(4a+1\right)$$