Question

explain

Let P(x) = x + ax4 + bx3 + cx? + dx +e. P(4) = P(5)=P(6)=P(7)=P(8) = 0. What is the value of a - b+c-d+e? Numerically the answer is somewhat large, but it can be represented easily as a sum of terms, where the terms are a product of 1 or more small integers. You may leave your answer in this form, instead of calculating it out.

Answer 1

For a polynomial equation of degree n ,

  $P(x)=x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+.................+a_{n}=o$ , whose roots are $\alpha _{1},\alpha _{2}....\alpha _{n}$

we have the relation between the roots and the coefficients as :

• the sum of roots =  $-a_{1}$
• the sum of products of roots taken two at a time =  $a_{2}$
• the sum of products of roots taken three at a time =  $-a_{3}$
• ..............................
• the sum of products of roots taken n at a time = $(-1)^{n}a_{n}$

Here we have

$P(x)=x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e$ where 4 , 5 , 6 , 7 , 8 are the roots of P(x)=0

Thus from the above mentioned relation , we have

  

$\Rightarrow a= (-4)+(-5)+(-6)+(-7)+(-8)$

$b=4*5+4*6+4*7+4*8+5*6+5*7+5*8+6*7+6*8+7*8$

$-c=4*5*6+4*5*7+4*5*8+4*6*7+4*6*8+4*7*8+5*6*7+5*6*8+5*7*8+6*7*8$

  

$d=4*5*6*7+4*5*6*8+4*6*7*8+4*5*7*8+5*6*7*8$

  $-e=4*5*6*7*8$

  

$\Rightarrow a-b+c-d+e=(-4)+(-5)+(-6)+(-7)+(-8)+(-4*5)+(-4*6)+(-4*7)+(-4*8)+(-5*6)+(-5*7)+(-5*8)+(-6*7)+(-6*8)+(-7*8)+(-4*5*6)+(-4*5*7)+(-4*5*8)+(-4*6*7)+(-4*6*8)+(-4*7*8)+(-5*6*7)+(-5*6*8)+(-5*7*8)+(-6*7*8)+(-4*5*6*7)+(-4*5*6*8)+(-4*6*7*8)+(-4*5*7*8)+(-5*6*7*8)+(-4*5*6*7*8)$