Tag: introduction to factorisation

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Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Factorise : $6xy^2 + 4x^2y$ 

  1. $2xy(3x+y)$
  2. $xy(3x+2y)$
  3. $2xy(2x+3y)$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The common factor between $6xy^2$ and $4x^2y$ is $2xy$ that is the HCF of $6xy^2$ and $4x^2y$ is $2xy$. 

Therefore, we take $2xy$ as a common factor in the expression $6xy^2+4x^2y$ as shown below:
$6xy^2+4x^2y=2xy(2x+3y)$
Hence, the factors of $6xy^2+4x^2y$ are $2xy$ and $(2x+3y)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Which of the following is an example of factorisation?

  1. $x^2+2x=x(x+2)$
  2. $x^2+2x=x(x+1)$
  3. $x^2+2x=x(x+3)$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The common factor between $x^2$ and $2x$ is $x$ that is the HCF of $x^2$ and $2x$ is $x$. 


Therefore, we take $x$ as a common factor in the expression $x^2+2x$ as shown below:


$x^2+2x=x(x+2)$


Hence, the factorization of $x^2+2x$ is $x(x+2)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Factorise : $5mn+15mnp$

  1. $5mn(1 + 3p)$
  2. $3mn(1 + 5p)$
  3. $5mn(1 - 3p)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The common factor between $5mn$ and $15mnp$ is $5mn$ that is the HCF of $5mn$ and $15mnp$ is $5mn$. 


Therefore, we take $5mn$ as a common factor in the expression $5mn+15mnp$ as shown below:


$5mn+15mnp=5mn(1+3p)$


Hence, the factorization of $5mn+15mnp$ is $5mn(1+3p)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Simplify: $\displaystyle \left( -80{ m }^{ 4 }npq \right) \div 10{ m }^{ 3 }{ pqn }^{ 2 }$

  1. $-8mn$
  2. $-8mnpq$
  3. $-8m$
  4. $\dfrac {-8m}{n}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \left( -80{ m }^{ 4 }npq \right) \div 10{ m }^{ 3 }{ pqn }^{ 2 }=\frac { -80{ m }^{ 4 }npq }{ 10{ m }^{ 3 }{ pqn }^{ 2 } } $

$\displaystyle =\quad -8\times \frac { { m }^{ 4 } }{ { m }^{ 3 } } \times \frac { n }{ { n }^{ 2 } } \times \frac { p }{ p } \times \frac { q }{ q } $
$\displaystyle =\quad -8{ m }^{ 4-3 }\times { n }^{ 1-2 }$
$\displaystyle =\quad -8m\times { n }^{ -1 }$
$\displaystyle =\quad \frac { -8m }{ n } \left( \because { n }^{ -1 }=\frac { 1 }{ n }  \right) $

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If $f(x)$ and $g(x)$ are two polynomials with integral coefficients which vanish at $x = \dfrac {1}{2}$, then what is the factor of HCF of $f(x)$ and $g(x)$?

  1. $x - 1$
  2. $x - 2$
  3. $2x - 1$
  4. $2x + 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $x = \dfrac {1}{2}$

$ \Rightarrow (2x - 1) = 0$
Therefore, $ (2x - 1)$ is satisfying both $f(x)$ and $g(x)$, so $(2x - 1)$ is the factor of H.C.F. of $f(x)$ and $g(x)$.

Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx+c=0, a \neq 0$ then the wrong statement is

  1. $\alpha ^2+\beta ^2=\dfrac{b^2-2ac}{a^2}$
  2. $\alpha \beta =\frac{c}{a}$
  3. $\alpha +\beta =\frac{b}{a}$
  4. $\frac{1}{\alpha }+\frac{1}{\beta }=-\frac{b}{c}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
We know that sum of roots $= - \dfrac{b}{a}$ and product of roots is  $\dfrac{c}{a}$
Hence option (C) is wrong statement.
Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

Consider the following statements :
1. $x - 2$ is a factor of $x^{3} - 3x^{2} + 4x - 4$
2. $x + 1$ is a factor of $2x^{3} + 4x + 6$
3. $x - 1$ is a factor of $x^{6} - x^{5} + x^{4} - x^{3} + x^{2} - x + 1$
Of these statements

  1. 1 and 2 are correct

  2. 1, 2 and 3 are correct

  3. 2 and 3 are correct

  4. 1 and 3 are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
  1. Remainder $=2^3-3\times 2^2+4\times 2-4$
    $= 8-12+8-4 = 0$
    Hence $x-2$ is a factor.
    2. Remainder$= 2(-1)^3+4(-1)+6$
    $= -2-4+6 = 0$
    Hence $x + 1$ is a factor.
    3. Ramainder $= 1^6-1^5+1^5-1^3+1^2-1+1 = 1$
    Hence $x - 1$ is not a factor.
    $\therefore$ Statements 1 and 2 are correct.