Question

factor completely. 5z² + 30z + 25

Explanation:

Step1: Identify a and c

For quadratic \(5z^2 + 30z + 25\), \(a = 5\), \(c = 25\). Find \(a\times c=5\times25 = 125\).

Step2: Find factors of 125

Find two numbers that multiply to 125 and add to 30 (the coefficient of \(z\)). The numbers are 25 and 5 (\(25\times5 = 125\), \(25 + 5 = 30\)).

Step3: Split the middle term

Rewrite the middle term: \(5z^2+25z + 5z + 25\).

Step4: Group and factor

Group the first two and last two terms: \((5z^2 + 25z)+(5z + 25)\). Factor out common factors: \(5z(z + 5)+5(z + 5)\). Then factor out \((z + 5)\): \((5z + 5)(z + 5)\). Factor out 5 from the first binomial: \(5(z + 1)(z + 5)\). Wait, no, wait. Wait, original quadratic: \(5z^2+30z + 25\). Wait, let's check again. Wait, \(a = 5\), \(b = 30\), \(c = 25\). Let's use the formula for factoring \(ax^2+bx + c\). First, factor out the GCF of 5: \(5(z^2+6z + 5)\). Now factor \(z^2+6z + 5\). Find two numbers that multiply to 5 and add to 6: 5 and 1. So \(z^2+6z + 5=(z + 5)(z + 1)\). So the complete factorization is \(5(z + 1)(z + 5)\)? Wait, no, wait, original quadratic: \(5z^2+30z + 25\). Let's check by expanding \(5(z + 5)(z + 1)\): \(5(z^2+6z + 5)=5z^2+30z + 25\). Yes, that's correct. Wait, but earlier when I split the middle term, I had \(5z^2+25z+5z + 25\), which is \(5z(z + 5)+5(z + 5)=(5z + 5)(z + 5)=5(z + 1)(z + 5)\). Yes, that's correct.

Wait, but wait, maybe I made a mistake. Wait, let's check the discriminant: \(b^2-4ac=30^2-4\times5\times25=900 - 500 = 400\). Square root of 400 is 20. So the roots are \(\frac{-30\pm20}{2\times5}=\frac{-30 + 20}{10}=\frac{-10}{10}=-1\), and \(\frac{-30 - 20}{10}=\frac{-50}{10}=-5\). So the quadratic factors as \(5(z + 1)(z + 5)\) (since the leading coefficient is 5, and the roots are -1 and -5, so \(a(x - r_1)(x - r_2)=5(z - (-1))(z - (-5))=5(z + 1)(z + 5)\)).

Wait, but let's do it again. Start with \(5z^2+30z + 25\). Factor out 5: \(5(z^2+6z + 5)\). Now factor \(z^2+6z + 5\). We need two numbers that multiply to 5 and add to 6. Those numbers are 5 and 1. So \(z^2+6z + 5=(z + 5)(z + 1)\). Therefore, the complete factorization is \(5(z + 1)(z + 5)\).

Wait, but let's check by multiplying back: \(5(z + 1)(z + 5)=5[(z)(z)+z(5)+1(z)+1(5)]=5[z^2+5z+z + 5]=5[z^2+6z + 5]=5z^2+30z + 25\), which matches the original quadratic. So that's correct.

Answer:

\(5(z + 1)(z + 5)\)