Answer:[tex]\displaystyle 24x^{2} - 41x + 12 = 24\left(x - \frac{3}{8}\right) \cdot \left(x - \frac{4}{3}\right) = (8x-3)\cdot (3x - 4)[/tex].Step-by-step explanation:Apply the quadratic formula to find all factors. For a quadratic equation in the form [tex]a\cdot x^{2} + b\cdot x + c = 0[/tex],where [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex] are constants, the two roots will be[tex]\displaystyle x_1 = \frac{-b + \sqrt{b^{2} - 4\cdot a \cdot c}}{2a}[/tex], and[tex]\displaystyle x_2 = \frac{-b - \sqrt{b^{2} - 4\cdot a \cdot c}}{2a}[/tex].For this quadratic polynomial, [tex]a = 24[/tex],[tex]b = -41[/tex], and[tex]c = 12[/tex].Apply the quadratic formula to find any [tex]x[/tex] value or values that will set this polynomial to zero:[tex]\displaystyle x_1 = \frac{-(-41) + \sqrt{(-41)^{2} - 4\times 24 \times 12}}{2\times 24} = \frac{3}{8}[/tex].[tex]\displaystyle x_2 = \frac{-(-41) - \sqrt{(-41)^{2} - 4\times 24 \times 12}}{2\times 24} = \frac{4}{3}[/tex].Apply the factor theorem to find the two factors of this polynomial:[tex]\displaystyle \left(x - \frac{3}{8}\right)[/tex] for the root [tex]\displaystyle x = \frac{3}{8}[/tex], and[tex]\displaystyle \left(x - \frac{4}{3}\right)[/tex] for the root [tex]\displaystyle x = \frac{4}{3}[/tex].Keep in mind that simply multiplying the two factors will not reproduce the original polynomial. Doing so assumes that the leading coefficient of [tex]x[/tex] in the original polynomial is one, which isn't the case for this question.Multiply the product of the two factors by the leading coefficient of [tex]x[/tex] in the original polynomial. [tex]\displaystyle 24\left(x - \frac{3}{8}\right) \cdot \left(x - \frac{4}{3}\right) = (8x-3)\cdot (3x - 4)[/tex].Expand to make sure that the factored form is equivalent to the original polynomial:[tex](8x-3)\cdot (3x - 4)\\ = (8\times 3)x^{2} + ((-3)\times 3 + (-4)\times 8)\cdot x + ((-3)\times (-4))\\ = 24x^{2} - 41x + 12[/tex].