One factor of x^3+2x^2-23x-60 is x+4. Find the remaining factors.
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One factor of x^3+2x^2-23x-60 is x+4. Find the remaining factors.

[From: ] [author: ] [Date: 11-04-24] [Hit: ]
. .-23. . . .......
Show work please

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By dividing x^3 + 2x^2 - 23x - 60 by x + 4 using Synthetic Division:
-4. . . .|. . . .1. . . .2. . . .-23. . . .-60
. . . . . . . . . . . . .-4. . . . .8. . . . .60
. . . . . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
. . . . . . . . .1. . . .-2. . . .-15. . . . .0

So:
(x^3 + 2x^2 - 23x - 60)/(x + 4) = x^2 - 2x - 15
==> x^3 + 2x^2 - 23x - 60 = (x + 4)(x^2 - 2x - 15).

Since x^2 - 2x - 15 factors to (x + 3)(x - 5), the final factorization is:
x^3 + 2x^2 - 23x - 60 = (x + 4)(x + 3)(x - 5).

I hope this helps!

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If (x+4) is a factor of x^3 + 2x^2 - 23x - 60. Then there must be a quadratic factor that can be multiplied by (x+4) to get x^3 + 2x^2 - 23x - 60. Remember a quadratic factor is of the form ax^2 + bx + c
Therefore:
x^3 + 2x^2 - 23x - 60 = (x+4)(ax^2 + bx + c)
x^3 + 2x^2 - 23x - 60 = ax^3 + bx^2 + cx + 4ax^2 + 4bx + 4c
x^3 + 2x^2 - 23x - 60 = ax^3 + (b + 4a)x^2 + (c + 4b)x + 4c
By inspection you can see that a = 1, b + 4a = 2, c + 4b = -23 and 4c = -60

as a = 1, then b + 4 = 2, therefore: b = -2
as b = -2, then c - 8 = -23 therefore: c = - 15
another way to determine c would be 4c = -60 therefore c = -15
that means the quadratic factor we are after is: ( x^2 - 2x - 15)
This now means that:
x^3 + 2x^2 - 23x - 60 = (x + 4)(x^2 - 2x - 15)

However (x^2 - 2x - 15) can still be further simplified. It can be divided into two linear factors:

Using the quadratic equation:
x = [ 2 ±√ ((-2)^2 - (4*-15)) ] / 2
x = (2 ±√64)/2
x = (2±8)/2
x = (2 + 8)/2 = 5
or:
x = (2 - 8)/2 = -3

Therefore the quadratic factor (x^2 - 2x - 15) ca be written as (x-5)(x+3)

So now
x^3 + 2x^2 - 23x - 60 = (x+4)(x-5)(x+3)

So the remaining factors were (x-5) and (x+3)

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Either perform synthetic division or simply factorise the polynomial as I have done below.

x^3+2x^2-23x-60
=x^3+4x^2-2x^2-8x-15x-60
=x^2(x+4)-2x(x+4)-15(x+4)
=(x+4)(x^2-2x-15)
=(x+4){x^2-(5-3)x-15}
=(x+4)(x^2-5x+3x-15)
=(x+4){x(x-5)+3(x-5)}
=(x+4)(x-5)(x+3)

Therefore, the other two factors are (x-5) and (x+3).

Hope this helps!!!

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Use Synthetic division

Answer (x+4)(x^2-2x-15)

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x^3+2x^2-23x-60 = (x+4)*(x^2-2x-15) =
(x+4)*(x-5)*(x+3)
1
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