Find the HCF and LCM of the polynomials.
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QUESTION -
x^3 - 9x^2 + 26 x - 24 and x^3 - 12 x ^ 2 + 47 x - 60
ANSWER -
x^3 - 12 x ^ 2 + 47 x - 60
------>
For convenience reduce the coefficient of x^2 to zero after setting equal to zero of the equation
x^3+9x^2+26x+24.
Put x = z+a {translation on x axis}
(z^3+ 3z^2 a+3z a^2+ a^3) + 9(z^2+2za+a^2)+26(z+a)+24
z^3+z^2(3a+9)+z(3a^2+9a+26)
+(a^3+9a^2+26a+24)=0.
It points to (aimed at making coefficient of z^2 equal to zero)
3a+9=0 => a= -3, substituting which
z^3 - 26z = 0.
z(z^2 -26) = 0.
one of the roots is z
=> z = x- a = x+3 is a root.
By long division,
x+3) x^3+9x^2+26x+24 (x^2+6x+8
. . . . x^3+3x^2
. . . . ________
. . . . . . +6x^2+26x
. . . . . . .+6x^2+18x
. . . . . . ._________
. . . . . . . . . . . + 8x+24
. . . . . . . . . . . + 8x+24
. . . . . . . . . . . ------------.
Hence it is factored first as
(x+3)(x^2+6x+8)
by inspection
x^2+6x+8 = (x+4)(x+2);
it is factored first as
it is factored first as(x+3)(x+2)(x+4).
For x^3 - 12 x ^ 2 + 47 x - 60
x³ - 12x² + 47x - 60 = 0
You are told a zero is 3. If that is true, then x - 3 has to be a factor. If it is, then you'll have a quadratic in return that you should be able to factor easy (or at least find its roots easy enough).
So we have:
. . . . _x²_-_9x_+_20______
x - 3 ) x³ - 12x² + 47x - 60
. . . . . x³ - 3x²
. . . . --------------
. . . . . . . . -9x² + 47x - 60
. . . . . . . . -9x² + 27x
. . . . . . . -----------------
. . . . . . . . . . . . 20x - 60
. . . . . . . . . . . . 20x - 60
. . . . . . . . . . . ---------------
. . . . . . . . . . . . . . . . . 0
And there is no remainder, so this shows that 3 is a root.
So now our equation is:
x³ - 12x² + 47x - 60 = 0
(x - 3)(x² - 9x + 20) = 0
And this should be easier to factor:
(x - 3)(x - 4)(x - 5) = 0
Your three roots are:
Your three roots are:x = 3, 4, and 5