# CLASS-8SQUARE OF TRINOMIAL

SQUARE OF A TRINOMIAL –

Let us find the square of the trinomial  x + y + z

So, (x + y + z)² = {(x + y) + z }²

= {(x + y)² + 2.(x + y).z + z²}

[Applying the formula (a+b)²= a²+2ab+b²]

=  (x² + 2xy + y²) + 2xz + 2yz + z²

[Applying the formula (a+b)²= a²+2ab+b²]

=  x²+ 2xy + y² + 2xz + 2yz + z²

= x²+ y²+ z² + 2(xy + yz + zx)

There are some example are given below for your better understanding –

Example.1)  (2a + 4b + 6c)²

Ans.)  (2a + 4b + 6c)²

=  {(2a + 4b) + 6c}²

=  (2a + 4b)²+ 2 X 6c X (2a + 4b) + 36c²

=  4a²+ 16b²+ 2 X 2aX 4b + 12c (2a + 4b) + 36c²

=  4a²+ 16b²+ 16ab + 24ac + 48bc + 36c²

=  4 (a²+ 4b²+ 9c²+ 4ab + 6ac + 12bc)              (Ans.)

Example.2)    (3x + 5y - 7z)²

Ans.)   (3x + 5y - 7z)²

=  {(3x + 5y) - 7z}²

=  (3x + 5y)²- 2 X (3x + 5y) X 7z + 49z²

=  9x²+ 2 X 3x X 5y + 25y²- 42xz – 70yz + 49z²

=  9x²+ 25y²+ 49z²+ 30xy – 42xz – 70yz            (Ans.)

Example.3)  (2p – 3q + 4r)²

Ans.)  (2p – 3q + 4r)²

=  {(2p – 3q) + 4r}²

=  {(2p – 3q)²+ 2 X (2p – 3q) X 4r + (4r)²}

= (4p²- 2 X 2p X 3q + 9q²) + 16pr – 24qr + 16r²

=  4p²- 12pq + 9q² + 16pr – 24qr + 16r²

=  4p² + 9q² + 16r² - 12pq + 16pr – 24qr        (Ans.)

Example.4)   (2a – 3x – 5p)²

Ans.)    (2a – 3x – 5p)²

=  {2a – (3x + 5p)}²

=  4a²- 2 X 2a X (3x + 5p) + (3x + 5p)²

=  4a²- 12ax – 20ap + ( 9x² + 2 X 3x X 5p + 25p²)

=  4a²- 12ax – 20ap + 9x²+ 30 xp + 25p²

=  4a²+ 9x²+ 25p²- 12ax – 20ap + 30xp        (Ans.)

OR,

(2a – 3x – 5p)²

=  {(2a – 3x) – 5p}²

=  (2a – 3x)²- 2 X (2a – 3x) X 5p + 25p²

=  4a²- 2 X 2a X 3x + 9x²- 20ap + 30xp + 25p²

=  4a²+ 9x²+ 25p²- 12ax – 20ap + 30xp          (Ans.)

Corollaries –

Two corollaries follow from this expansion

1) (a + b + c)² - 2(ab + bc + ca)

=  a² + b² + c² + 2(ab + bc + ca) - 2(ab + bc + ca)

=  a² + b² + c²

So,   a²+ b²+ c² = (a + b + c)² - 2(ab + bc + ca)

2)  (a + b + c)² - (a² + b² + c²)

= a²+ b²+ c²+ 2(ab + bc + ca) - (a² + b² + c²)

=  2(ab + bc + ca)

So,  2(ab + bc + ca) = (a + b + c)² - (a² + b² + c²)

There are some examples are given below for your better understanding -

Example.1) If a + b + c = 10 and ab + bc + ca = 20, find a²+ b²+ c²

Ans.)  As per the given condition, a + b + c = 10 and ab + bc + ca = 20, we have to find out the value of a²+ b²+ c² = ?

As per the formula,  2(ab + bc + ca) = (a + b + c)² - (a² + b² + c²)

So,  (a² + b² + c²) =  (a + b + c)² - 2(ab + bc + ca)

=  10² - (2X20)  (putting the value)

=  100 – 40  =  60

So,  (a²+ b²+ c²) =  60              (Ans.)

Example.2)  If, a + b + c = 15 and a²+ b²+ c² = 125, then find ab + bc + ca = ?

Ans.) As per the given condition, a + b + c = 15 and a²+ b²+ c² = 125, we have to find out the value of ab + bc + ca = ?

As per the formula,  2(ab + bc + ca) =  (a + b + c)² - (a² + b² + c²)

2(ab + bc + ca) = 15² - 125    (putting the value)

2(ab + bc + ca) =  225 - 125 =  100

ab + bc + ca  =  100/2  =  50          (Ans.)