Question

Factories 24x^2-41x+12

Answer 1

Answer:

$\displaystyle 24x^{2} - 41x + 12 = 24\left(x - \frac{3}{8}\right) \cdot \left(x - \frac{4}{3}\right) = (8x-3)\cdot (3x - 4)$.

Step-by-step explanation:

Apply the quadratic formula to find all factors. For a quadratic equation in the form

$a\cdot x^{2} + b\cdot x + c = 0$,

where $a$, $b$, and $c$ are constants, the two roots will be

$\displaystyle x_1 = \frac{-b + \sqrt{b^{2} - 4\cdot a \cdot c}}{2a}$, and

$\displaystyle x_2 = \frac{-b - \sqrt{b^{2} - 4\cdot a \cdot c}}{2a}$.

For this quadratic polynomial,

• $a = 24$,
• $b = -41$, and
• $c = 12$.

Apply the quadratic formula to find any $x$ value or values that will set this polynomial to zero:

$\displaystyle x_1 = \frac{-(-41) + \sqrt{(-41)^{2} - 4\times 24 \times 12}}{2\times 24} = \frac{3}{8}$.

$\displaystyle x_2 = \frac{-(-41) - \sqrt{(-41)^{2} - 4\times 24 \times 12}}{2\times 24} = \frac{4}{3}$.

Apply the factor theorem to find the two factors of this polynomial:

• $\displaystyle \left(x - \frac{3}{8}\right)$ for the root $\displaystyle x = \frac{3}{8}$, and
• $\displaystyle \left(x - \frac{4}{3}\right)$ for the root $\displaystyle x = \frac{4}{3}$.

Keep in mind that simply multiplying the two factors will not reproduce the original polynomial. Doing so assumes that the leading coefficient of $x$ 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 $x$ in the original polynomial.

$\displaystyle 24\left(x - \frac{3}{8}\right) \cdot \left(x - \frac{4}{3}\right) = (8x-3)\cdot (3x - 4)$.

Expand to make sure that the factored form is equivalent to the original polynomial:

$(8x-3)\cdot (3x - 4)\\ = (8\times 3)x^{2} + ((-3)\times 3 + (-4)\times 8)\cdot x + ((-3)\times (-4))\\ = 24x^{2} - 41x + 12$.