How do you factor 24x^2 + 4x - 21?

Question

I'm unable to factor this problem. Is this problem factorable?

Answer 1

It is not factorable in integers. The roots of the quadratic ax^2 + bx + c are found by the formula x = (-b +- sqrt(b^2 - 4ac) / 2a, which for this quadratic simplifies to (-1 +- sqrt(127)) / 12.

For the quadratic to be factorable in integers, the roots need to be either integers, or nice fractions. For example, if one root was exactly 3/7, the quadratic would have the factor (7x - 3). This will only happen if (b^2 - 4ac) is a perfect square. But the roots of this quadratic are IRRATIONAL, because of sqrt(127) - or sqrt(2032) before simplifying, as noted by a previous answerer. Their values are 0.85578564 and -1.022452306.

Answer 2

The factors are:
[x - (-1-127^0.5)/12] and [x - (-1+127^0.5)/12] or
[x + 1.0225] and [x - 0.8558]

To factorise this expression, you can use the solution formula for quadratic equation, i.e.
x = [-b +/- (b^2 - 4ac)^0.5]/(2a)

This is because the discriminant b^2 - 4ac is not a perfect square.

Alternatively, you may use the perfecting square method:
24x^2 + 4x - 21
= 24[x^2 + 1/6x - 21/24]
= 24[x^2 + 1/6x + (1/12)^2 - 21/24 - (1/12)^2]
= 24[(x + 1/12)^2 - 127/144]
= 24(x + 1/12)^2 - 127/6

Let the expression equal to zero, and solve for x. You will get:
x = -1.0225 and x = 0.8558
Thus, the factors are (x + 1.0225) and (x - 0.8558)...
same as above.

Answer 3

Factoring is not possible

The quadratic formula, the discriminate √b² - 4ac

√(4)² - 4(24)(21)

√16 - 2016

√- 2000

contains a negative number and no solution.

Tried the perfect square without success.

- - - - - - - - - - - -s-

Answer 4

$ mathomatic
Mathomatic version 12.5.10 (www.mathomatic.org)
Copyright (C) 1987-2006 George Gesslein II.
50 equation spaces available, 960000 bytes per equation space.

1-> 24x^2 + 4x - 21

#1: (24*(x^2)) + (4*x) - 21

1-> factor

#1: (4*x*((6*x) + 1)) - 21

Answer 5

what's f(x + a million) if f(x) = 6x3 - 3x2 + 4x - 9? a. 6x3 + 12x2 + 4x + 2 b. 6x3 + 3x2 + 8x + 6 c. 6x3 + 21x2 + 20x + 4 d. 6x3 + 15x2 + 16x - 2 For the 1st difficulty, replace x with (x+a million).

Answer 6

There is no way that you can multiple two numbers to get the third term and add together to get the middle term. You also can't pull out any factors because there are no common ones. I'm pretty good at math and I can't figure it out. Sometimes there just aren't any solutions.

Hope this helps.

Answer 7

You will need the quadratic formula

24x^2 + 4x - 21
x=(-4+/-sqrt(16+4*24*21))/48=-1/12+/-(1/48)sqrt(16+2016)
x=-1/12+/-(1/48)sqrt(2032)--1/12+/-(1/24)sqrt(127)
factors
(x+1/12+(1/24)sqrt(127))(x+1/12-(1/24)sqrt(127))

Answer 8

The factor is not in integer form, just like gopal did up there

Answer 9

Find the roots x1 and x2 using the quadratic formula. If the roots are real then you can factor the expression as:

(x - x1)(x - x2)

Answer 10

it is not factorable since 2032 is not power of some integer number.