Solve x^2-41x+400

$${ x }^{ 2 } -41x+400$$

Factor

$$\left(x-25\right)\left(x-16\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+400. To find a and b, set up a system to be solved.

$$a+b=-41 ab=1\times 400=400$$

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 400.

$$-1,-400 -2,-200 -4,-100 -5,-80 -8,-50 -10,-40 -16,-25 -20,-20$$

Calculate the sum for each pair.

$$-1-400=-401 -2-200=-202 -4-100=-104 -5-80=-85 -8-50=-58 -10-40=-50 -16-25=-41 -20-20=-40$$

The solution is the pair that gives sum -41.

$$a=-25 b=-16$$

Rewrite x^{2}-41x+400 as \left(x^{2}-25x\right)+\left(-16x+400\right).

$$\left(x^{2}-25x\right)+\left(-16x+400\right)$$

Factor out x in the first and -16 in the second group.

$$x\left(x-25\right)-16\left(x-25\right)$$

Factor out common term x-25 by using distributive property.

$$\left(x-25\right)\left(x-16\right)$$

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.

$$x^{2}-41x+400=0$$

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.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 400}}{2}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 400}}{2}$$

Multiply -4 times 400.

$$x=\frac{-\left(-41\right)±\sqrt{1681-1600}}{2}$$

Add 1681 to -1600.

$$x=\frac{-\left(-41\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-41\right)±9}{2}$$

The opposite of -41 is 41.

$$x=\frac{41±9}{2}$$

Now solve the equation x=\frac{41±9}{2} when ± is plus. Add 41 to 9.

$$x=\frac{50}{2}$$

Divide 50 by 2.

$$x=25$$

Now solve the equation x=\frac{41±9}{2} when ± is minus. Subtract 9 from 41.

$$x=\frac{32}{2}$$

Divide 32 by 2.

$$x=16$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 25 for x_{1} and 16 for x_{2}.

$$x^{2}-41x+400=\left(x-25\right)\left(x-16\right)$$

Evaluate

$$\left(x-25\right)\left(x-16\right)$$