Question

The polynomial f(x) = x⁴ - 2x³ - px² + q is divisible by (x-2)². Find the values p and q. Hence, solve the equation.

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{The\;polynomial\;f(x)=x^4-2x^3-px^2+q\;is\;divisible\;by\;(x-2)^2}$

$\underline{\textsf{To find:}}$

$\textsf{The values of p and q}$

$\underline{\textsf{Solution:}}$

$\mathsf{Since\;f(x)\;is\;divisible\;by\;(x-2)^2,\,we\;can\;write}$

$\mathsf{x^4-2x^3-px^2+q=(x^2-4x+4)(x^2+ax+\frac{q}{4})}$

$\mathsf{Equating\;coefficient\;of\;x^3\;on\;bothsies,}$

$\mathsf{-2=a-4}$

$\implies\boxed{\mathsf{a=2}}$

$\mathsf{Equating\;coefficient\;of\;x^2\;on\;bothsies,}$

$\mathsf{-p=4+\frac{q}{4}-4a}$

$\mathsf{-p=4+\frac{q}{4}-8}$

$\mathsf{-p=\frac{q}{4}-4}$

$\mathsf{-4p=q-16}$

$\mathsf{4p+q=16}$.......(1)

$\mathsf{Equating\;coefficient\;of\;x\;on\;bothsies,}$

$\mathsf{0=-q+4a}$

$\mathsf{0=-q+8}$

$\implies\boxed{\mathsf{q=8}}$

$\mathsf{(1)\implies\;4p+8=16}$

$\mathsf{4p=8}$

$\implies\boxed{\mathsf{p=2}}$

$\mathsf{Consider,}$

$\mathsf{x^2+2x+2=0}$

$\mathsf{x^2+2x+1=-1}$

$\mathsf{(x+1)^2=i^2}$

$\mathsf{x+1=\pm\,i}$

$\implies\mathsf{x=-1\pm\,i}$

$\mathsf{(x-2)^2=0\implies\;x=2,2}$

$\underline{\textsf{Answer:}}$

$\mathsf{Roots\;of\;f(x)\;are\;2,2,-1+i,-1-i}$

$\underline{\textsf{Find more:}}$

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Answer 2

SOLUTION

GIVEN

The polynomial f(x) = x⁴ - 2x³ - px² + q is divisible by (x-2)²

TO DETERMINE

• The values p and q

• To solve the equation

CONCEPT TO BE IMPLEMENTED

If c is a multiple root of the polynomial equation f(x) = 0 of order r then c is a multiple root of the equation f' (x) = 0 of order ( r - 1 )

EVALUATION

Here it is given that

f(x) = 4x³ - 6x² - 2px

∴ f ' (x) = x⁴ - 2x³ - px² + q

It is also given that f(x) is divisible by (x-2)²

So 2 is a multiple root of f(x) of order 2

∴ f(2) = 0 & f ' (2) = 0

f(2) = 0 gives

16 - 16 - 4p + q = 0

$\sf{q = 4p} \: \: \: \: \: \: .......(1)$

Again f ' (2) = 0 gives

32 - 24 - 4p = 0

$\implies \sf{4p = 8}$

$\implies \sf{p = 2}$

From Equation (1) we get

$\sf{q = 4 \times 2}$

∴ q = 8

So the given equation becomes

f(x) = x⁴ - 2x³ - px² + q

∴ f(x) = x⁴ - 2x³ - 2x² + 8

Now we factorise it as below :

$\sf{f(x) = {x}^{4} - 2 {x}^{3} - 2 {x}^{2} + 8 }$

$\sf{ = {x}^{3} (x - 2 ) - 2( {x}^{2} - 4) }$

$\sf{ = (x - 2 )( {x}^{3} - 2x - 4) }$

$\sf{ = (x - 2 )( {x}^{3} - 2 {x }^{2} + 2 {x}^{2} - 4x + 2x - 4) }$

$\sf{ = (x - 2 )(x - 2)( {x}^{2} + 2x + 2) }$

$\sf{ = {(x - 2)}^{2} ( {x}^{2} + 2x + 2) }$

Now (x - 1)² = 0 gives x = 1 , 1

Again

$\sf{( {x}^{2} + 2x + 2)} = 0 \: \: gives$

$\displaystyle \sf{x = \frac{ - 2 \: \pm \: \sqrt{4 - 4.1.2} }{2.1} }$

$\displaystyle \sf{ = \frac{ - 2 \: \pm \: \sqrt{ - 4 } }{2} }$

$\displaystyle \sf{ = \frac{ - 2 \: \pm \: 2i }{2} }$

$= \sf{ -1 \: \pm \: i}$

So the roots are

$\sf{2 \: , \: 2 \: , \: - 1 + i \: , \: - 1 - i}$

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