We have, a = kl , b = (lm +kn) and c = mn, where ac is the product of kl and mn that is, equal to the product of lm and kn which are the coefficient of x. Therefore (kl × mn) = (lm ×kn) .

**Factorising the Quadratic
Polynomial (Trinomial) of the type ax^{2} **

The
linear factors of *ax*^{2} + *bx* +*c* will be in the form (*kx* +
*m*) and (*lx* + *n*)

Thus,
*ax*^{2} + *bx* +*c* = (*kx* +*m* )(*lx* +*n*) = *klx*^{2} +(*lm* + *kn* )*x* +*mn*

Comparing
coefficients of *x*^{2}, *x* and constant term *c* on both
sides.

We have,
*a* = *kl* , *b* = (*lm* +*kn*) and *c* = *mn*, where *ac*
is the product of kl and *mn* that is, equal to the product of *lm* and
*kn* which are the coefficient of *x*. Therefore (*kl* ×
*mn*) = (*lm* ×*kn*) .

Step 1 :
Multiply the coefficient of *x*^{2} and constant term, that is *ac*
.

Step 2 :
Split ac into two numbers whose sum and product is equal to *b* and *ac*
respectively.

Step
3 : The terms are grouped into two pairs and factorise.

Factorise
2*x*^{2} + 15*x* +
27

*Solution*

Compare with
*ax*^{2} + *bx* +*c*

we get, *a*
=
2, *b* = 15, *c* = 27

product
*ac* = 2 ×27 = 54 and sum *b* =
15

We
find the pair 6, 9 only satisfies “*b* = 15” and also “*ac* = 54”.

we
split the middle term as 6*x* and 9*x*

2*x*
^{2} + 15*x* + 27 = 2*x* ^{2} +
6*x* + 9*x* + 27

= 2*x*
(*x* + 3) + 9(*x* +
3)

=
(*x* + 3)(2*x* + 9)

Therefore,
(*x* + 3) and (2*x* + 9) are the factors of 2*x* ^{2}
+
15*x* + 27 .

Factorise
2*x*^{2} − 15*x* +
27

Compare
with *ax* ^{2} + *bx* +*c*

*a
*=*
*2,* b *= −*
*15,* c *=*
*27

product
*ac* = 2×27 = 54, sum *b*=–15

we
split the middle term as –6*x* and –9*x*

2*x*^{2}
−
15*x* + 27 = 2*x*^{2} −6*x*
−
9*x* + 27

= 2*x*(*x* − 3) − 9(*x* − 3)

= (*x* − 3)(2*x* −9)

Therefore,
(*x* − 3) and (2*x* − 9) are the factors of 2*x*^{2} −
15*x* + 27.

Factorise
2*x*^{2} + 15*x* −27

Compare
with *ax* ^{2} + *bx* +*c*

Here,
*a* = 2, *b* = 15, *c* = −27

product
*ac* = 2×–27 = –54, sum *b*=15

we
split the middle term as 18*x* and –3*x*

2*x*^{2}
+
15*x* −27 = 2*x*^{2} +18*x*
−
3*x* −27

= 2*x*(*x* + 9) − 3(*x* + 9)

= (*x* + 9)(2*x* − 3)

Therefore,
(*x* + 9) and (2*x* − 3) are the factors of 2*x*^{2}
+
15*x* −27.

Factorise
2*x*^{2} − 15*x* -27

Compare
with *ax* ^{2} + *bx* +*c*

Here,*a*
=
2, *b* = − 15, *c* = −27

product
*ac* = 2×–27=–54, sum *b*=–15

we
split the middle term as –18*x* and 3*x*

2*x*
^{2} − 15*x* -27 = 2*x* ^{2} −18*x* + 3*x* −27

=
2*x* (*x* −9) + 3(*x* −9)

=
(*x* −9)(2*x* + 3)

Therefore,
(*x* − 9) and (2*x* + 3) are the factors of 2*x*^{2}
− 15*x* -27

Factorise
(*x* + *y* )^{2} + 9(*x *+ *y* ) + 20

Let
*x* + *y* = *p* , we get *p*^{2}
+
9*p* + 20

Compare
with *ax*^{2} + *bx* +*c*,

We
get *a* = 1, *b* = 9, *c* =
20

product
*ac* = 1×20 = 20, sum *b*=9

we
split the middle term as 4*p* and 5*p*

*p*^{2}* *+* *9* p *+*
*20* *=*
p*^{2}* *+* *4*p *+*
*5* p *+*
*20

*=
p*(*p *+*
*4)* *+*
*5(*p *+*
*4)

=
( *p* + 4)(*p* + 5)

Put,
*p* = *x* +*y* we get, (*x* + *y* )^{2} +
9(*x* + *y* ) + 20 = (*x* +*y* + 4)(*x* +*y* + 5)

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9th Maths : UNIT 3 : Algebra : Factorising the Quadratic Polynomial (Trinomial) | Steps, Example Solved Problems | Algebra | Maths

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