Question

Factories 24x^2-41x+12

Answer 1

`\displaystyle 24x^(2) - 41x + 12 = 24\left(x - (3)/(8)\right) \cdot \left(x - (4)/(3)\right) = (8x-3)\cdot (3x - 4)`.

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* 24 * 12}}{2* 24} = (3)/(8)`.

`\displaystyle x_2 = \frac{-(-41) - \sqrt{(-41)^(2) - 4* 24 * 12}}{2* 24} = (4)/(3)`.

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

• 
  `\displaystyle \left(x - (3)/(8)\right)` for the root
  `\displaystyle x = (3)/(8)`, and
• 
  `\displaystyle \left(x - (4)/(3)\right)` for the root
  `\displaystyle x = (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 - (3)/(8)\right) \cdot \left(x - (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* 3)x^(2) + ((-3)* 3 + (-4)* 8)\cdot x + ((-3)* (-4))\\ = 24x^(2) - 41x + 12`.