Question

factor each trinomial if the trinomial cannot be factored write prime. show ALL work1.) 2x^2+5x-122.) 6x^2-23x+7

Answer 1

1) the initial expression is:

$2x^2+5x-12$

To factor this expression we can use the cuadratic formula that is:

$\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4(a)(c)}}{2(a)} \\ a=2 \\ b=5 \\ c=-12 \end{gathered}$

so we replace the values and we get:

$\begin{gathered} x=\frac{-5\pm\sqrt[]{5^2-4(2)(-12)}}{2\cdot2} \\ x=\frac{-5\pm\sqrt[]{121}}{4} \\ x=\frac{-5\pm11}{4} \end{gathered}$

So now we have two solutions so:

$\begin{gathered} x_1=\frac{-5+11}{4}=\frac{6}{4}=\frac{3}{2}=1.5 \\ x_2=\frac{-5-11}{4}=-\frac{16}{4}=-4 \end{gathered}$

so the factor expression will be:

$(x-\frac{3}{2})(x+4)$

2) the initial expression is:

$6x^2-23x+7$

so we repeat the same procedure so:

$\begin{gathered} x=\frac{23\pm\sqrt[]{23^2-4(6)(7)}}{2(6)} \\ x=\frac{23\pm\sqrt[]{529-168}}{12} \\ x=\frac{23\pm\sqrt[]{361}}{12} \\ x=\frac{23\pm19}{12} \end{gathered}$

So the two possible solutions are:

$\begin{gathered} x_1=\frac{23+19}{12}=\frac{42}{12}=\frac{7}{2}=3.5 \\ x_2=\frac{23-19}{12}=\frac{4}{12}=\frac{2}{6}=\frac{1}{3}=0.33 \end{gathered}$

so the factor form is:

$(x-\frac{7}{2})(x-\frac{1}{3})$