Olympiad Test : Algebra - 1 - Class 6 MCQ

Let the expression to be subtracted be X.
(a³ − 4a² + 5a − 6)-X = (a² − 2a + 1)
X = (a³ − 4a² + 5a − 6)- (a² − 2a + 1)
Since '-' sign precedes the parenthesis, we remove it and change the sign of each term within the parenthesis.
X = a³ − 4a² + 5a − 6- a² + 2a - 1
Rearranging and collecting the like terms:
X = a³ +(− 4-1)a² + (5+2)a − 6 - 1
X = a³ −5a² + 7a − 7
So, a³ −5a² + 7a − 7 must be subtracted from a³ − 4a² + 5a − 6 to obtain a² − 2a + 1.

 Required difference = 1– (2x – 3y – 4)
= 3y – 2x + 5

Required difference
= (l – 2m + 3n) – (2l – 4m – n)
= – l + 2m + 4n

Required value
= x² + y² – z²
= (–2)² + (–1)²– (3)² = 4 + 1 – 9
= 5 – 9 = – 4

In the expression -7l² mn, the coefficient of a variable is the constant factor that multiplies that variable.

The term involving m is -7l² mn. The coefficient of m is everything multiplying m, which in this case is -7l² n.

Thus, the correct coefficient of m is -7l² n.

2x² –y² + 3z²
= 2(4)² – (–1)² + 3(–2)²
= 32 – 1 + 12 = 44 – 1= 43

x³ + 3x² – x + 1 – 5x³ + 2x² – 6x – 7
= –4 x³ + 5x² – 7x – 6

Required difference
= – (–2x – 5y + 6z – 2)
= – 8y + 10z – 2

2x –[3y–{2x– (y–x)}]
= 2x –[3y–{2x– y + x)}]
= 2x –[3y–3x + y] = 2x – 4y + 3x = 5x–4y