Solve 25-20r+4r^2

$$25 - 20 r + 4 r ^ { 2 }$$

Factor

$$\left(2r-5\right)^{2}$$

Solution Steps

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

$$4r^{2}-20r+25$$

Factor the expression by grouping. First, the expression needs to be rewritten as 4r^{2}+ar+br+25. To find a and b, set up a system to be solved.

$$a+b=-20 ab=4\times 25=100$$

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 100.

$$-1,-100 -2,-50 -4,-25 -5,-20 -10,-10$$

Calculate the sum for each pair.

$$-1-100=-101 -2-50=-52 -4-25=-29 -5-20=-25 -10-10=-20$$

The solution is the pair that gives sum -20.

$$a=-10 b=-10$$

Rewrite 4r^{2}-20r+25 as \left(4r^{2}-10r\right)+\left(-10r+25\right).

$$\left(4r^{2}-10r\right)+\left(-10r+25\right)$$

Factor out 2r in the first and -5 in the second group.

$$2r\left(2r-5\right)-5\left(2r-5\right)$$

Factor out common term 2r-5 by using distributive property.

$$\left(2r-5\right)\left(2r-5\right)$$

Rewrite as a binomial square.

$$\left(2r-5\right)^{2}$$

Steps Using Square Of Binomial

This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.

$$factor(4r^{2}-20r+25)$$

Find the greatest common factor of the coefficients.

$$gcf(4,-20,25)=1$$

Find the square root of the leading term, 4r^{2}.

$$\sqrt{4r^{2}}=2r$$

Find the square root of the trailing term, 25.

$$\sqrt{25}=5$$

The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.

$$\left(2r-5\right)^{2}$$

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.

$$4r^{2}-20r+25=0$$

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.

$$r=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 25}}{2\times 4}$$

Square -20.

$$r=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 25}}{2\times 4}$$

Multiply -4 times 4.

$$r=\frac{-\left(-20\right)±\sqrt{400-16\times 25}}{2\times 4}$$

Multiply -16 times 25.

$$r=\frac{-\left(-20\right)±\sqrt{400-400}}{2\times 4}$$

Add 400 to -400.

$$r=\frac{-\left(-20\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$r=\frac{-\left(-20\right)±0}{2\times 4}$$

The opposite of -20 is 20.

$$r=\frac{20±0}{2\times 4}$$

Multiply 2 times 4.

$$r=\frac{20±0}{8}$$

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

$$4r^{2}-20r+25=4\left(r-\frac{5}{2}\right)\left(r-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4r^{2}-20r+25=4\times \frac{2r-5}{2}\left(r-\frac{5}{2}\right)$$

Subtract \frac{5}{2} from r by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4r^{2}-20r+25=4\times \frac{2r-5}{2}\times \frac{2r-5}{2}$$

Multiply \frac{2r-5}{2} times \frac{2r-5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4r^{2}-20r+25=4\times \frac{\left(2r-5\right)\left(2r-5\right)}{2\times 2}$$

Multiply 2 times 2.

$$4r^{2}-20r+25=4\times \frac{\left(2r-5\right)\left(2r-5\right)}{4}$$

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

$$4r^{2}-20r+25=\left(2r-5\right)\left(2r-5\right)$$

Evaluate

$$\left(2r-5\right)^{2}$$