Answer:
\(4x^{2} +23x-72=(4x-9)(x+8)\)
Step-by-step explanation:
we have
\(4x^{2} +23x-72\)
Equate the quadratic equation to zero
\(4x^{2} +23x-72=0\)
Group terms that contain the same variable, and move the constant to the opposite side of the equation
\(4x^{2} +23x=72\)
Factor the leading coefficient
\(4(x^{2} +(23x/4))=72\)
Complete the square. Remember to balance the equation by adding the same constants to each side
\(4(x^{2} +(23x/4)+(529/64))=72+(529/16)\)
\(4(x^{2} +(23x/4)+(529/64))=1681/16\)
\((x^{2} +(23x/4)+(529/64))=1681/64\)
Rewrite as perfect squares
\((x+(23/8))^{2}=1681/64\)
square root both sides
\((x+(23/8))=(+/-)41/8\)
\(x=-(23/8)(+/-)41/8\)
\(x=-(23/8)+41/8=18/8=9/4\)
\(x=-(23/8)-41/8=-64/8=-8\)
therefore
\(4x^{2} +23x-72=4(x-9/4)(x+8)\)
\(4x^{2} +23x-72=(4x-9)(x+8)\)
