Method of undetermined coefficients for PreRMO, PRMO and IITJEE Foundation maths

  1. Find out when the expression x^{3}+px^{2}+qx+r is exactly divisible by x^{2}+ax+b

Solution 1:

Let x^{3}+px^{2}+qx+r=(x^{2}+ax+b)(Ax+B) where A and B are to be determined in terms of p, q, r, a and b. We can assume so because we know from the fundamental theorem of algebra that the if the LHS has to be of degree three in x, the remaining factor in RHS has to be linear in x.

So, expanding out the RHS of above, we get:



We are saying that the above is true for all values of x: hence, coefficients of like powers of x on LHS and RHS are same; we equate them and get a system of equations:





Hence, we get p=a+\frac{r}{b} and bp-ba=r or that b(p-a)=r

Also, b+aB=q so that q=b+\frac{ar}{b} which means q-b=\frac{a}{b}r

but \frac{r}{b}=B=p-a and hence, q-b=\frac{a}{b}(p-a)

So, the required conditions are b(p-a)=r and q-b=\frac{a}{b}(p-a).

2) Find the condition that x^{2}+px+q may be a perfect square.

Solution 2:

Let x^{2}+px+q=(Ax+B)^{2} where A and B are to be determined in terms of p and q; finally, we obtain the relationship required between p and q for the above requirement.

x^{2}+px+q=A^{2}x^{2}+B^{2}+2ABx which is true for all real values of x;

Hence, A^{2}=1 so A=1 or A=-1

Also, B^{2}=q and hence, B=\sqrt{q} or B=-\sqrt{q}

Also, 2AB=p so that 2\sqrt{q}=p so q=\frac{p^{2}}{4}, which is the required condition.

3) To prove that x^{4}+px^{3}+qx^{2}+rx+s is a perfect square if (q-\frac{p^{2}}{4})^{2}=4s and r^{2}=p^{2}s.

Proof 3:

Let x^{4}+px^{3}+qx^{2}+rx+s=(Ax^{2}+Bx+C)^{2}







A=1 or A=-1

2AB=p \longrightarrow 2B=p \longrightarrow B=\frac{p}{2}

q=B^{2}+2AC=\frac{p^{2}}{4}+2\times \sqrt{s} \longrightarrow (q-\frac{p^{2}}{4})^{2}=4s

2 \times \frac{p}{2} \times \sqrt{s}=r \longrightarrow r^{2}=p^{2}s

More later,

Nalin Pithwa.

PS: Note in the method of undetermined coefficients, we create an identity expression which is true for all real values of x.

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