In this section, we use the synthetic division method that helps to factorise a cubic polynomial into linear factors.

__Factorisation using Synthetic
Division__

In this
section, we use the synthetic division method that helps to factorise a cubic polynomial
into linear factors. If we identify one linear factor of cubic polynomial *p*(*x*)
then using synthetic division we can get the quadratic factor of *p*(*x*).
Further if possible one can factorise the quadratic factor into linear factors.

**Note**

**•**** **For any non constant polynomial** ***p*(*x*)*, x = a*** **is zero if and only if** ***p*(*a*) = 0

**•**** ***x–a *is a factor for* p*(*x*)*
*if and only if* p*(*a*) = 0* *(Factor theorem)

**To identify (***x*** – **1**) and (***x*** + 1) are the factors of a polynomial**

**•**** **(*x*–1) is a factor
of *p*(*x*) if and only if the sum of coefficients of *p*(*x*)
is 0.

**•**** **(*x*+1) is a factor
of *p*(*x*) if and only if the sum of the coefficients of even power of
*x,* including constant is equal to the sum of the coefficients of odd powers
of *x*

**Example 3.38**

(i)
Prove that (*x* -1) is a factor of *x* ^{3}
−
7*x*^{2} +13*x* −7

(ii)
Prove that (*x* +1) is a factor of *x* ^{3}
+
7*x*^{2} +13*x* +
7

*Solution*

(i)
Let *p*(*x*) = *x* ^{3} −7*x*
^{2} + 13*x* −7

Sum
of coefficients = 1 −7 + 13 −7 = 0

Thus
(*x* -1) is a factor of *p*(*x*)

(ii)
Let *q*(*x*) = *x* ^{3} +
7*x* ^{2} + 13*x* +
7

Sum
of coefficients of even powers of *x* and constant term = 7 + 7 = 14

Sum
of coefficients of odd powers of *x* = 1 + 13 = 14

Hence,
(*x* +1) is a factor of *q*(*x*)

Factorise
*x* ^{3} + 13*x*^{2} +
32*x* + 20 into linear factors.

Let, *p*(*x*)* *=* x *^{3}* *+13*x *^{2}* *+*
*32*x *+*
*20

Sum
of all the coefficients = 1 +13 + 32 + 20 = 66 ≠ 0

Hence,
(*x* -1) is not a factor.

Sum
of coefficients of even powers and constant term = 13 + 20 = 33

Sum
of coefficients of odd powers = 1 + 32 = 33

Hence,
(*x* +1) is a factor of *p*(*x*)

Now
we use synthetic division to find the other factors

**Example 3.40**

Factorise
*x* ^{3} − 5*x* ^{2} −2*x*
+
24

Let
*p*(*x* ) = *x* ^{3} −5*x*
^{2} − 2*x* + 24

When
*x* = 1, *p*(1)= 1 −5 – 2 +24 =18 ≠ 0 (*x* -1)is not a factor.

When
*x* = –1, *p*(-1)= −1 – 5+2+24
=20≠0 (*x* +1) is not a factor.

Therefore,
we have to search for different values of *x* by trial and error method.

When
*x* = 2

*p*(2)* *=* *2^{3}* *−5(2)^{2}* *−*
*2(2)* *+*
*24

= 8−20−4+24

= 8 ≠ 0 Hence, (*x*-2) is not a
factor

When
*x* = − 2

*p*(-2)=* *(−2)^{3}* *−* *5(−2)^{2}* *−*
*2(−2)* *+*
*24

= −8−20+4+24

*p*(-2)* *=* *0

Hence,
(*x*+2) is a factor

Thus,
(*x* + 2)(*x* − 3)(*x* −
4) are the factors.

Therefore,
*x* ^{3} − 5*x*^{2} −2*x*
^{2} + 24 = (*x* +
2)(*x* − 3)(*x* − 4)

Note

Check whether 3 is a zero of *x*2* *−* *7*x *+* *12* *. If it is not, then* *check for –3 or 4 or –4 and so on.

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