Q. 32

# Solve the followi

In factorization, we write the middle term of the quadratic equation either as a sum of two numbers or difference of two numbers such that the equation can be factorized.

⇒ 6(x2 + 12 – 7x + x2 + 4 – 5x + x2 – 3x + 2) = (x2 – 3x + 2)(x2 + 12 – 7x)
⇒ 6(3x2 + 18 – 15x ) = x4 + 12x2 – 7x3 – 3x3 – 36x + 21x2 + 2x2 + 24 – 14x

⇒ 18x2 – 90x + 108 = x4 + 12x2 – 7x3 – 3x3 – 36x + 21x2 + 2x2 + 24 – 14x

⇒ x4 – 10x3 + 17x2 + 40x + 84 = 0

Let P(x) = x4 – 10x3 + 17x2 + 40x + 84
At x = -2,
(-2)4 - 10(-2)3 + 17(-2)2 + 40(-2) + 84 = 16 + 80 + 68 - 80 + 84
P(x) = 0
therefore, x + 2 is a factor of P(x).
On dividing P(x) by (x + 2), we get x3 - 12x2 + 41x - 42
Let g(x) = x3 - 12x2 + 41x - 42, P(x) = (x - 2)g(x)
at x =-2
g(x) =0
therefore, x + 2 is a factor of g(x).
On dividing g(x) by (x + 2), we get x2 - 14x + 49
Therefore,
P(x) = (x - 2)(x - 2)(x2 - 14x + 49)
Using, (a - b)2 =a2 + b2 - 2ab, we have
P(x) =  (x + 2)2(x – 7)2

⇒ (x + 2)2(x – 7)2 = 0
Therefore, possible value of 'x' are -2, -2, and -7, -7

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