Solve 4x^2-5+x=

$$4 { x }^{ 2 } -5+x=$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$4x^{2}+x-5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=1 ab=4\left(-5\right)=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. 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 1.

$$a=-4 b=5$$

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

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

Factor out 4x in the first and 5 in the second group.

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

Factor out common term x-1 by using distributive property.

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

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.

$$4x^{2}+x-5=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-5\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-1±\sqrt{1-16\left(-5\right)}}{2\times 4}$$

Multiply -16 times -5.

$$x=\frac{-1±\sqrt{1+80}}{2\times 4}$$

Add 1 to 80.

$$x=\frac{-1±\sqrt{81}}{2\times 4}$$

Take the square root of 81.

$$x=\frac{-1±9}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-1±9}{8}$$

Now solve the equation x=\frac{-1±9}{8} when ± is plus. Add -1 to 9.

$$x=\frac{8}{8}$$

Divide 8 by 8.

$$x=1$$

Now solve the equation x=\frac{-1±9}{8} when ± is minus. Subtract 9 from -1.

$$x=-\frac{10}{8}$$

Reduce the fraction \frac{-10}{8} to lowest terms by extracting and canceling out 2.

$$x=-\frac{5}{4}$$

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}{4} for x_{2}.

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

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

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

Add \frac{5}{4} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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

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

Evaluate

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