# (x-1)(x-2)(2x-1)(2x+1)-70=0

2
by Pyu
which syllabus are these questions? are these in secondary school?

2014-10-18T21:15:45+05:30
(x-1)(x-2(((2x-1)(2x+1))-70=0
here (2x-1)(2x+1) is in the form of (a+b)(a-b)
formulae for it is a^2-b^2
here a=2x and b=1
substituting the values we get
(2x-1)(2x+1)=(2x)^2-(1)^2
=4x^2-1
(x-1)(x-2)=x(x-2)-1(x-2)
=x^2-2x-x+2
=x^2-x+2
(x-1)(x-2)(2x-1)(2x+1)-70=0
(x^2-x+2)(4x^2-1)-70=0
By simplification we get
4x^4-12x^3+7x^2+3x-72=0
By long division method we can write as
(x-3)(4x^3+7x+24)=0
Again by long division method we can write as
(x-3)(2x+3)(2x^2-3x+8)=
by using the principle of zero products
x-3=0 or 2x+3=0 or 2x^2-3x+8=0
Thus the values are
x-3=0
x=3
2x+3=0
2x=-3
x=-3/2
2x^2+3x+2=0
By simplification
x=0.7500+1.8540i and x=0.7500-1.8540i
U r welcome
2014-10-19T05:02:29+05:30

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The rational factors, if they exist, are one or some of the following :
+-  factors of 72 / factors of 4
=  +- 1, 2,3,4, 6,8,9,12,18,36,72,   +- 1/2, 1/4, 3/2, 3/4, 9/2,  9/4

On checking them you find that  x = 3 and x = -3/2  satisfy the polynomial.

Let us rewrite the polynomial in the following way. and find the value of "a"

(x-3) (x+3/2) 4 (x² + a x - 72/(4*-3*3/2) ) = 0
(x² - 3/2 x - 9/2) 4 (x² + a x + 4) = 0
Coefficient of x³ will be : 4 a - 4 * 3/2 = - 12
a = -3/2

So   x² - 3/2 x + 4 = 0

So there are two real and two imaginary roots.

================================
Trying the rational roots:

P(0) = -ve   P(4) = 308 = +ve.  so a root exists between x= 0 & 4.
Trying   P(3) =0   x = 3 is a root.

P(-4) = 1820 = +ve   So a root exists between 0 and -4.
Trying  -1, -2, and -3/2  works.