Olympiad Test: Algebraic Expressions - Class 7 MCQ

The constant term in the expression 1+x²+x is 1.

Coefficient is the numerical number associated with the variable. So here we have (-1)y³. So the answer is -1.

multiplication of a number with variable gives multiplication of integer with integer and variables with variables

in given question, it'll lead to

2a x 3a = 6a²

To find the product of (2x+3y)(2x+3y), you can use the distributive property

(2x + 3y)(2x + 3y)

= (2x)(2x) + (2x)(3y) + (3y)(2x) + (3y)(3y)

= 4x² + 6xy + 6xy + 9y²

Now, combine like terms:

= 4x² + 12xy + 9y²

So, the product of (2x+3y)(2x+3y) is 4x² + 12xy + 9y²

To find the degree of the polynomial:

x³ - x²y² - 8y² + 2

We need to determine the highest sum of the exponents in each term:

• x³: The degree is 3.
• -x²y²: The degree is (2 + 2 = 4).
• -8y²: The degree is 2.
• Constant term 2: The degree is 0 (constants have degree 0).

The highest degree among these terms is 4, which comes from the term -x²y².

Therefore, the degree of the polynomial is: (b) 4

3x - 5a -x 2 + 9b = 3*3 - 5(-1)-32 + 9 (-2) = 9 + 5 -9 - 18 = 5 - 18 = -13

2b-8x+4x²+4a
=2(-2)-8(3)+ 4(3)²+4(-1)
=-4-24+36-4
=4

3a – 2b – ab – (a – b + ab) + 3ab + b – a
write similar terms with each other
3a-a-a-2b+b+b-ab-ab+3ab
a+ab

y²-x²-(x²-y²)=y²-x²-x²+y²=2y²-2x²=-2(x²-y²)

3x² - 4y² + 5xy + 20-  (-x² - y² + 6xy+20)
=3x² - 4y² + 5xy + 20 + x² + y² - 6xy-20)
=4x²-3y²-xy

Calculation:

Given: f(x,y,z) = x y̅ z + x̅ y̅ z + x y̅ z̅ 

x y̅ z + x y̅ z̅ + x̅ y̅ z

x y̅ ( z + z̅) + x̅ y̅ z

x y̅ + x̅ y̅ z

y̅ (x + x̅ z)

y̅ (x + x̅ )(x + z)

y̅ z + y̅ x