Can someone factor 6x3 - 23x2 - 5x plus 4 completely given that 2x plus 1 is a factor?

Answer 1

The factors are of the form (ax + b)(cx +d)(ex + f) then, the coefficient of x³ is the product of the coefficients of the three x terms, namely ace = 6

The coefficient of x² is the sum of the product of the coefficients of x from 2 brackets multiplied by the number from the remaining bracket.

The coefficient of x is the sum of the product of the numbers from two brackets multiplied by the x coefficient from the remaining bracket.

The final number is the product of the numbers from the three brackets giving bdf = 4.

Some number crunching gives the solution (3x - 1)(x - 4)(2x + 1)

Alternatively:

We are told that 2x + 1 is a factor. Thus we know:

(2x + 1)(ax² + bx + c) = 6x³ - 23x² - 5x + 4

Multiplying out the left gives:

2ax³ + (2b + a)x² + (2c + b)x + c = 6x³ - 23x² - 5x + 4

Thus: 2a = 6 (coefficients of x³) → a = 3

And 2b + a = 2b + 3 = -23 (coefficients of x²) → b = -13

And c = 4 (term not including any x).

Thus we now know;

6x³ - 23x² - 5x + 4 = (2x +1)(3x² - 13x + 4)

As we are told to factorise it completely, it is likely that the quadratic also factorises. As 3 only has the factor pair 1×3, the quadratic will factorise into (3x + d)(x + e)

Thus 3x³ + (3e + d)x + de = 3x² -13x + 4

As the term without any x is positive, d and e must have the same sign.

As the coefficient of the x term is negative, both d and e must be negative.

Thus, the factor pairs of 4 to consider are (-1)×(-4) and (-2)×(-2).

As the coefficient of x is -13 = 3e + d, the values e = -4 and -1 are [fairly] obvious.

Thus 6x³ - 23x² - 5x + 4 = (2x +1)(3x - 1)(x - 4)