Test: Factorisation- 1 - Class 8 MCQ

∴  5(2x + 1)(3x + 5) (2x + 1) = 5(3x+5)

26xy(x + 5) (y - 4) / 13x(y - 4)
Cancelling 13x(y-4)
=2y(x+5)

By the distributive property a(b-c)=ab- ac
so 4(x-5)=4x-20

x (3x + 2)= 3x²+2x (by distributive property a(b+c)=ab+ac)

10x² − 18x³ + 14x⁴ = 2x(5x− 9x² + 7x³) = 2x²(5− 9x + 7x²)

x^2 + xy + 8x + 8y = x[x + y] + 8[x + y]

 = (x + y)(x + 8)

We have 4y² - 12y+9. comparing the equation with (a-b)²=a²-2ab+b²,gives us a²=(2y)²,2ab=2*3*2y and b²=(3)².Hence the answer is (2y-3)².

7x²y²z²÷14xyz
=(7/14)(x²/x)(y²/y)(z²/z)
=1/2(x)(y)(z)
=1/2xyz.

Only 2y is common in both the terms

Factors are numbers we can multiply together to get another number. A constituent or element that brings about certain effects or results, or indicates a specific multiple, number, or quantity.

x⁴- (y+z)⁴, applying x² - y²=(x-a)(x+a), gives (x²-(y+z)²)(x²+(y+z)²). Again applying the same identity gives (x-(y+z))(x+(y+z))(x²+(y+z)²)

x⁴-(x-z)⁴ = (x²-(x-z)²)(x²+(x-z)²) = (x-(x-z))(x+(x-z)(x²+(x-z)²) = z(2x-z)(x²+(x-z)²(Applying a² - b² = (a+b)(a-b))

8 (x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z² 

= 

(x³+2x²+3x)/2x
=x(x²+2x+3)/2x
=(x²+2x+3)/2

LHS = 2x + 3y
RHS = 5xy
LHS ≠ RHS
Correct statement would be
2x + 3y = 2x+3y

An equation is a mathematical statement with an 'equal to' symbol between two algebraic expressions that have equal values.

Given,

x + 2x + 3x = 5x

Solving LHS we get,

x + 2x + 3x

= x(1 + 2 + 3) = 6x [By adding them, as they are like terms]

But RHS = 5x

Thus, L.H .S. ≠ R.H.S.

The correct statement is x + 2x + 3x = 6x

Let us take 12 as common
and then the factor is 12(x+3) .

15xy - 6x + 5y - 2 = 3x[5y-2] + 1[5y-2]
= (5y - 2)(3x + 1)