Question

solving quadratic equations by factoring. solve each equation by factoring. 1) (k + 1)(k - 5)=0 2) (a + 1)(a + 2)=0 3) (4k + 5)(k + 1)=0 4) (2m + 3)(4m + 3)=0

Explanation:

Step1: Apply zero - product property for (1)

If \((k + 1)(k - 5)=0\), then either \(k+1 = 0\) or \(k - 5=0\).
For \(k+1 = 0\), we get \(k=-1\); for \(k - 5=0\), we get \(k = 5\).

Step2: Apply zero - product property for (2)

If \((a + 1)(a + 2)=0\), then either \(a+1 = 0\) or \(a + 2=0\).
For \(a+1 = 0\), we get \(a=-1\); for \(a + 2=0\), we get \(a=-2\).

Step3: Apply zero - product property for (3)

If \((4k + 5)(k + 1)=0\), then either \(4k+5 = 0\) or \(k + 1=0\).
For \(4k+5 = 0\), \(4k=-5\), so \(k=-\frac{5}{4}\); for \(k + 1=0\), \(k=-1\).

Step4: Apply zero - product property for (4)

If \((2m + 3)(4m+3)=0\), then either \(2m + 3=0\) or \(4m+3=0\).
For \(2m + 3=0\), \(2m=-3\), so \(m=-\frac{3}{2}\); for \(4m+3=0\), \(4m=-3\), so \(m=-\frac{3}{4}\).

Answer:

(1) \(k=-1,k = 5\)
(2) \(a=-1,a=-2\)
(3) \(k=-\frac{5}{4},k=-1\)
(4) \(m=-\frac{3}{2},m=-\frac{3}{4}\)