Solve (4x^2-21x+5)

$$( 4 x ^ { 2 } - 21 x + 5 )$$

Factor

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

Solution Steps

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.

$$a+b=-21 ab=4\times 5=20$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 20.

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

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -21.

$$a=-20 b=-1$$

Rewrite 4x^{2}-21x+5 as \left(4x^{2}-20x\right)+\left(-x+5\right).

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

Factor out 4x in the first and -1 in the second group.

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

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

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

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.

$$4x^{2}-21x+5=0$$

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.

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

Square -21.

$$x=\frac{-\left(-21\right)±\sqrt{441-4\times 4\times 5}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-21\right)±\sqrt{441-16\times 5}}{2\times 4}$$

Multiply -16 times 5.

$$x=\frac{-\left(-21\right)±\sqrt{441-80}}{2\times 4}$$

Add 441 to -80.

$$x=\frac{-\left(-21\right)±\sqrt{361}}{2\times 4}$$

Take the square root of 361.

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

The opposite of -21 is 21.

$$x=\frac{21±19}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{21±19}{8}$$

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

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

Divide 40 by 8.

$$x=5$$

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

$$x=\frac{2}{8}$$

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

$$x=\frac{1}{4}$$

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}.

$$4x^{2}-21x+5=4\left(x-5\right)\left(x-\frac{1}{4}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-21x+5=4\left(x-5\right)\times \frac{4x-1}{4}$$

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

$$4x^{2}-21x+5=\left(x-5\right)\left(4x-1\right)$$

Evaluate

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