Now we discuss a few additional information with which we can solve higher degree polynomials.

**Polynomials with Additional Information**

Now we discuss a few additional information with which we can
solve higher degree polynomials. Sometimes the additional information will
directly be given, like, one root is 2 +
3*i *. Sometimes the additional information like, sum of the coefficients
is zero, have to be found by observation of the polynomial.

If *α** *+
*i**β** *is an imaginary root of
a quartic polynomial with real coefficients, then *α** *− *i**β** *is also a root;
thus (*x *−
(*α** *+ *i**β** *)) and (*x *− (*α** *− *i**β** *)) are factors of the
polynomial; hence their product is a factor; in other words, *x*^{2} − 2*α**x *+ *α*^{2} + *β*^{2} is a factor; we can
divide the polynomial with this factor and get the second degree quotient which
can be solved by known techniques; using this we can find all the roots of the
polynomial.

If 2+ √3 is a root of a quadric polynomial equation with rational coefficients,
then 2 - √3 is also a root; thus
their product (*x*-(2+√3))
(*x*-(2-√3)) is a factor; that is *x*^{2} − 4*x *+1 is a factor; we can
divide the polynomial with this factor and get the quotient as a second degree
factor which can be solved by known techniques. Using this, we can find all the
roots of the quadric equation. This technique is applicable for all surds taken in
place of 2 + √3.

If an imaginary root and a surd root of a sixth degree
polynomial with rational coefficient are known, then step by step we may reduce
the problem of solving the sixth degree polynomial equation into a problem of
solving a quadratic equation.

If 2 + i and 3 - √2 are roots of the equation

*x*^{6} -13*x*^{5} + 62*x*^{4} -126*x*^{3} + 65*x*^{2} +127*x *-140 = 0
,

find all roots.

Since the coefficient of the equations are all rational numbers,
and 2 + i and 3 - √2 are roots, we get 2 - i and 3 + √2 are also roots of the
given equation. Thus (x - (2 + *i*)), (x - (2 - *i*)), (x - (3 - √2))
and (x - (3 + √2)) are factors. Thus their product

((x - (2 + i))(x - (2 - i))(x - (3 - √2))(x - (3 + √2))

is a factor of the given polynomial equation. That is,

(*x*^{2} - 4*x *+ 5)(*x*^{2} - 6*x *+ 7)

is a factor. Dividing the given polynomial equation by this
factor, we get the other factor as (*x*^{2} - 3*x *- 4) which implies that 4 and -1
are the other two roots. Thus

2 + *i*, 2 - *i*, 3 + √2, 3 -√2, -1, and 4

are the roots of the given polynomial equation.

If *P*(*x*) is a polynomial equation of degree
2*n *, having only even powers of *x *, (that is,
coefficients of odd powers are 0 ) then by replacing *x*^{2} by *y
*, we get a polynomial equation with degree *n *in *y*;
let *y*_{1} , *y*_{2} ,... *y*_{n}* *be the roots of this polynomial equation. Then
considering the *n *equations *x*^{2} = *yr *, we can find two values for *x *for each *y*_{r}* *; these 2_{n}* *numbers are the roots
of the given polynomial equation in *x *.

**Example 3.16**

Solve the equation *x*^{4} - 9*x*^{2} + 20 = 0 .

**Solution**

The given equation is

*x*^{4} - 9*x*^{2} + 20 = 0 .

This is a fourth degree equation. If we replace *x*^{2} by *y *, then we
get the quadratic equation

*y*2 - 9 *y *+ 20 = 0 .

It is easy to see that 4 and 5 as solutions for y^{2} - 9 y + 20 = 0 . Now
taking *x*^{2} = 4 and x^{2} = 5 , we get 2, -2, √5, -√5 as solutions of the given
equation.

We note that
the technique adopted
above can be
applied to polynomial equations like *x*^{6} -17*x*^{3} + 30 = 0 , *ax*^{2k}* *+ *bx*^{k}* *+ *c *= 0 and in
general polynomial equations of the form *a*_{n}* x*^{kn}* *+ *a*_{n}_{-1} *x*^{k }^{(n-1)} +...+ *a*_{1} *x*^{k}* *+ *a*_{0} = 0 where *k *is
any positive integer.

Let *P *(
*x*) = 0 be a polynomial
equation such that the sum of the coefficients is zero. What actually the sum
of coefficients is? The sum of coefficients is nothing but *P*(1). The sum
of all coefficients is zero means that *P*(1) = 0 which says that 1 is
a root of *P*(*x*) . The rest of the problem of solving the equation
is easy.

**Example 3.17**

Solve the equation *x*^{3} - 3*x*^{2} - 33*x *+ 35 = 0
.

**Solution**

The sum of the coefficients of the polynomial is 0. Hence 1 is a
root of the polynomial. To find other roots, we divide *x*^{3} - 3*x*^{2} - 33*x *+ 35 by *x
*-1 and get *x*^{2} - 2*x *– 35 as the quotient. Solving this we get 7 and -5
as roots. Thus 1, 7, -5 form the solution set of the given equation.

Let *P *(
*x*) = 0 be a polynomial
equation such that the sum of the coefficients of the odd powers and that of
the even powers are equal. What does actually this mean? If *a *is the
coefficient of an odd degree in *P *(
*x*) = 0 , then the
coefficient of the same odd degree in *P *(−*x*)
= 0 is −*a *. The coefficients of
even degree terms of both *P*(*x*) = 0 and *P*(−*x*) =
0 are same. Thus the given condition implies that the sum of all coefficients
of *P*(−*x*) = 0 is zero and hence 1
is a root of *P*(−*x*) =
0 which says that −1 is a root of *P*(*x*) = 0 . The rest of the problem of solving the
equation is easy.

Solve the equation 2*x*^{3} +11*x*^{2} - 9*x *-18 = 0.

We observe that the sum of the coefficients of the odd powers
and that of the even powers are equal. Hence -1 is a root of the equation. To
find other roots, we divide 2*x*^{3} +11*x*^{2} - 9*x *-18 by *x
*+1 and get 2*x*^{2} + 9*x *-18 as the quotient. Solving this we get 3/2 and -6
as roots. Thus -6, -1, 3/2 are the roots or solutions of the given equation.

As already noted to solve higher degree polynomial equations, we
need some information about the solutions of the equation or about the
polynomial. “The roots are in arithmetic progression” and “the roots are in
geometric progression” are some of such information. Let us discuss an equation
of this type.

**Example 3.19**

Obtain the condition that the roots of *x*^{3} + *px*^{2} + *qx *+ *r *=
0 are in A.P.

**Solution**

Let the roots be in A.P. Then, we can assume them in the form *α
*- *d *,*α *,*α *+ *d *.

Applying the Vieta’s formula (α - d ) + α + (α + d ) = - p/1 = p ⇒
3α = - p ⇒ α =- p/3 .

But, we note that *α *is a root of the given equation.
Therefore, we get

**Example 3.20**

Find the condition that the roots of *ax*^{3} + *bx*^{2} + *cx *+ *d *=
0 are in geometric progression. Assume *a*, *b*, *c*, *d ≠ *0

**Solution**

Let the roots be in G.P.

Then, we can assume them in the form α/λ ,α ,α λ.

Applying the Vieta’s formula, we get

If the roots of *x*^{3} + *px*^{2} + *qx *+ *r *=
0 are in H.P. , prove that 9 *pqr *= 27*r*^{3} + 2*p*.

Assume *p*, *q*, *r *≠ 0

Let the roots be in H.P. Then, their reciprocals are in A.P. and
roots of the equation

Since the roots of (1) are in A.P., we can assume them as *α *-
*d *,*α *,*α *+ *d *.

Applying the Vieta’s formula, we get

Σ_{1} = (α - d ) + α + (α + d ) = - q/r ⇒ 3α = - q/r ⇒
α =- q/3r .

But, we note that *α *is a root of (1). Therefore, we get

**Example 3.22**

It is known that the roots of the equation *x*^{3} - 6*x*^{2} - 4*x *+ 24 = 0
are in arithmetic progression. Find its roots.

Let the roots be *a *- *d *, *a*, *a *+ *d
*. Then the sum of the roots is 3*a *which is equal to 6 from the given equation. Thus 3*a
*= 6 and hence *a *= 2 . The product of the roots is *a*^{3} - *ad*^{2} which is equal to -24
from the given equation. Substituting the value of *a *, we get 8 - 2*d *^{2} = -24 and hence *d *=
±4.

If we take *d *= 4 we get -2, 2, 6 as roots and if we take *d
*= -4, we get 6, 2, -2 as roots (same roots given in reverse order) of
the equation.

Tags : Solved Example Problems | Theory of Equations , 12th Mathematics : UNIT 3 : Theory of Equations

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