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

If the polynomial $f(x)$ is such that $f(-43) = 0$, which of the following is the factor of $f(x)$?

  1. $x - 43$
  2. $x$
  3. $x - 7$
  4. $x + 43$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f(-43)=0$ 
$\Rightarrow -43$ is root of $f(x)$

$\Rightarrow (x+43).g(x)=f(x)$ for some function $ g(x)$
$\Rightarrow (x+43)$ is the factor of $f(x)$

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

If the sum of two integers is $-2$ and their product is $-24$, the numbers are

  1. $6$ and $4$
  2. $-6$ and $4$
  3. $-6$ and $-4$
  4. $6$ and $-4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let $p$ and $q$ be the required integers

If $p+q$ and $pq$ are known then quadratic equation corresponding to roots as $p$ and $q$ is given by,
$x^2-(p+q)x+pq=0$
$\Rightarrow x^2+2x-24=0$, substitute the given values
$\Rightarrow x^2+6x-4x-24=0$, split the middle term
$\Rightarrow (x^2+6x)+(-4x-24)=0$, group pair of terms
$\Rightarrow x(x+6)-4(x+6)=0$, factor each binomials 
$\Rightarrow (x+6)(x-4)=0$, factor out common factor 
$\Rightarrow x=-6$ or $x=4$, set each factor to $0$

Hence $p$ and $q$ are $-6$ and $4$
Multiple choice maths algebraic formulae - expansion of squares introduction to factorization introduction to factorisation factorising algebraic expressions

The value of  $k$  for which  $x - 1$  is a factor of the polynomial  $4 x ^ { 3 } + 3 x ^ { 2 } - 4 x + k$  is

  1. $3$
  2. $0$
  3. $1$
  4. $-3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$x-1$ is a factor of $4{x^2} + 3{x^2} - 4x + k$


put $x=1$


$4{x^2} + 3{x^2} - 4x + k=0$

$ \Rightarrow 4{\left( 1 \right)^2} + 3{\left( 1 \right)^2} - 4\left( 1 \right) + k = 0$

$ \Rightarrow 4 + 3 - 4 + k = 0$

$ \Rightarrow k =  - 3$

Hence,
option $(D)$ is correct answer.

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) $