Factor

\left(z-5\right)^{2}

Evaluate

\left(z-5\right)^{2}

Quiz

Polynomial5 problems similar to: z ^ { 2 } - 10 z + 25## Similar Problems from Web Search

## Share

a+b=-10 ab=1\times 25=25

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

-1,-25 -5,-5

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 25.

-1-25=-26 -5-5=-10

Calculate the sum for each pair.

a=-5 b=-5

The solution is the pair that gives sum -10.

\left(z^{2}-5z\right)+\left(-5z+25\right)

Rewrite z^{2}-10z+25 as \left(z^{2}-5z\right)+\left(-5z+25\right).

z\left(z-5\right)-5\left(z-5\right)

Factor out z in the first and -5 in the second group.

\left(z-5\right)\left(z-5\right)

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

\left(z-5\right)^{2}

Rewrite as a binomial square.

factor(z^{2}-10z+25)

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.

\sqrt{25}=5

Find the square root of the trailing term, 25.

\left(z-5\right)^{2}

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.

z^{2}-10z+25=0

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.

z=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}

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.

z=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}

Square -10.

z=\frac{-\left(-10\right)±\sqrt{100-100}}{2}

Multiply -4 times 25.

z=\frac{-\left(-10\right)±\sqrt{0}}{2}

Add 100 to -100.

z=\frac{-\left(-10\right)±0}{2}

Take the square root of 0.

z=\frac{10±0}{2}

The opposite of -10 is 10.

z^{2}-10z+25=\left(z-5\right)\left(z-5\right)

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 5 for x_{2}.

x ^ 2 -10x +25 = 0

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

r + s = 10 rs = 25

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

r = 5 - u s = 5 + u

Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

(5 - u) (5 + u) = 25

To solve for unknown quantity u, substitute these in the product equation rs = 25

25 - u^2 = 25

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = 25-25 = 0

Simplify the expression by subtracting 25 on both sides

u^2 = 0 u = 0

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

r = s = 5

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.