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Question
The polynomial f(x) = ax4 + x3 + bx2 - 4x + c has (x + 1), (x-2) and (2x - 1) as its factors. Find the values of a,b,c, and remaining factor.
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Solution
When x + 1 is a factor, we can substitute x= -1 to evaluate values .... (i)
When x - 2 is a factor, we can substitute x = 2 to evaluate values .... (ii)
When 2x - 1 is a factor, we can substitute x = `1/2` to evaluate values ....(iii)
Substituting (1), we get
f(-1) = a × (-1)4 + (-1)3 + b(-1)2 - 4(-1) + c = 0
⇒ a + b + c = -3
⇒ a = - b - c - 3 ....(iv)
Substituting (11 ), we get
⇒ f(2) = a × (2)4 + (2)3 + b(2)2 - 4(2) + c = O
⇒ 16 a + 4 b + c = 0 .....(v)
Substituting (iii), we get
⇒ `"f"(1/2) = "a" xx (1/2)^4 + (1/2)^3 + "b"(1/2)^2 - 4(1/2) + c = 0`
`=> "a"/16 + "b"/4 + "c" = 2 -1/8`
⇒ a + 4b + 16c = 30 ..(vi)
Putting (iv) in (v) and (vi), we get:
16a + 4b + c = O; ⇒ 16x(- b - c - 3) + 4b+ c = O
⇒ -12 b - 15 c - 48 = 0 = 4b + 5c = 16
⇒ b = - 4 - `"5c"/4` .......(vii)
a + 4b + 16c = 30 ⇒ (- b- c - 3) + 4b + 16c = 30
⇒ 3b + 15 c = 33 .... (viii)
Putting (vii) 1n (viii) , we get,
⇒ 3 x `(-4 - (5"c")/4) + 15"c" = 33`,
⇒ Solving this , we get
⇒ c = 4
Putting this value of c in (viii), we get:
b = -9
Putting this value of c in (iv), we get:
a = 2
Puting values of a,b,c in polynomial , we get :
f (x) = 2x4 + x3 - 9x2 - 4x + 4
In order to find the remaining factor, lets start with (x-2) as one of the factor. Then,
Multiplying (x - 2) by 2x3, we get 2x4 - 4x3, hence we are left with 5x3 -9x2 - 4x + 4 (and 1st part of factor as 2x3).
Multiplying (x-2) by 5x2, we get 5x3 - 10x2, hence we are left with x2 - 4x + 4 (and 2"d part of factor as 5x2).
Multiplying (x-2) by x, we get x2 - 2x, hence we are left with -2x + 4 (and 3'd part of factor as x).
Multiplying (x - 2) by -2, we get -2x + 4, hence we are left with 0 (and 4th part of factor as -2).
Hence complete factor is (x - 2)(2x3 + 5x2 + x - 2).
Again factoring (2x3 + 5x2 + x - 2) by similar method we get factors as:
(x +1 )(2x - 1)(x + 2)
Hence, f(x) = 2x4 + x3 - 9x2 - 4x +4 = (x - 2)(x +1)(2x - 1)(x + 2).
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