Solve 10n^2+100n+250

$$10 { n }^{ 2 } +100n+250$$

Factor

$$10\left(n+5\right)^{2}$$

Solution Steps

Factor out 10.

$$10\left(n^{2}+10n+25\right)$$

Consider n^{2}+10n+25. Use the perfect square formula, a^{2}+2ab+b^{2}=\left(a+b\right)^{2}, where a=n and b=5.

$$\left(n+5\right)^{2}$$

Rewrite the complete factored expression.

$$10\left(n+5\right)^{2}$$

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.

$$factor(10n^{2}+100n+250)$$

Find the greatest common factor of the coefficients.

$$gcf(10,100,250)=10$$

Factor out 10.

$$10\left(n^{2}+10n+25\right)$$

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.

$$10\left(n+5\right)^{2}$$

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.

$$10n^{2}+100n+250=0$$

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.

$$n=\frac{-100±\sqrt{100^{2}-4\times 10\times 250}}{2\times 10}$$

Square 100.

$$n=\frac{-100±\sqrt{10000-4\times 10\times 250}}{2\times 10}$$

Multiply -4 times 10.

$$n=\frac{-100±\sqrt{10000-40\times 250}}{2\times 10}$$

Multiply -40 times 250.

$$n=\frac{-100±\sqrt{10000-10000}}{2\times 10}$$

Add 10000 to -10000.

$$n=\frac{-100±\sqrt{0}}{2\times 10}$$

Take the square root of 0.

$$n=\frac{-100±0}{2\times 10}$$

Multiply 2 times 10.

$$n=\frac{-100±0}{20}$$

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}.

$$10n^{2}+100n+250=10\left(n-\left(-5\right)\right)\left(n-\left(-5\right)\right)$$

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

$$10n^{2}+100n+250=10\left(n+5\right)\left(n+5\right)$$

Evaluate

$$10\left(n+5\right)^{2}$$