Question

Factor. v2 + 22v - 23

Answer 1

`v^2+22v-23`

we can factor this quadratic expression finding the zeros, as follows:

`\begin{gathered} v_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ v_(1,2)=\frac{-22\pm\sqrt[]{22^2-4\cdot1\cdot(-23)}}{2\cdot1} \\ v_(1,2)=\frac{-22\pm\sqrt[]{484+92}}{2} \\ v_(1,2)=\frac{-22\pm\sqrt[]{576}}{2} \\ v_1=(-22+24)/(2)=1 \\ v_2=(-22-24)/(2)=-23 \end{gathered}`

Then, the quadratic expression can be expressed as

`\begin{gathered} a(v-v_1)(v-v_2) \\ v^2+22v-23=(v-1)(v+23) \end{gathered}`