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

If $4x^{4} -12x^{3}+x^{^{2}}+3ax-b$ is divided by $x^{2}-1$ then a = ____, and b=___

  1. $5, 4$
  2. $4,9$
  3. $4,5$
  4. $1, -1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ x }^{ 2 }-1=0\quad \Rightarrow x=\pm 1$

Given a polynomial $P(x)={ 4x }^{ 4 }-12{ x }^{ 3 }+{ x }^{ 2 }+3ax-b$
Using remainder theorem,as $P(x)$ is completely divisible by ${ x }^{ 2 }-1$
$\therefore P(\pm 1)=0\ \therefore P(1)=0\ \Rightarrow 4-12+1+3a-b=0\ \Rightarrow 3a-b=7\ \therefore P(-1)=0\ \Rightarrow 4+12+1-3a-b=0\ \Rightarrow 3a+b=17$
By soling both we get
$a=4$   ;$b=5$

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

$7+3x$ is a factor of $3x^3+7x$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $7 + 3x$ is a factor of $p\left( x \right) = 3{x^3} + 7x$, then, $p\left( {\frac{{ - 7}}{3}} \right) = 0$.

Compute $p\left( {\frac{{ - 7}}{3}} \right)$ in the given polynomial.

$p\left( {\frac{{ - 7}}{3}} \right) = 3{\left( { - \frac{7}{3}} \right)^3} + 7\left( {\frac{{ - 7}}{3}} \right)$

$ =  - \frac{{343}}{9} - \frac{{49}}{3}$

$ = \frac{{ - 343 - 147}}{9}$

$ =  - \frac{{490}}{9}$

The value of $p\left( {\frac{{ - 7}}{3}} \right)$is not equal to 0.

The given statement is false.

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

Divide : $\displaystyle \left( 51{ m }^{ 3 }{ p }^{ 2 }-34{ m }^{ 2 }{ p }^{ 3 } \right)$ by $17mp$

  1. $\displaystyle { m }^{ 2 }p$
  2. $\displaystyle m{ p }^{ 2 }$
  3. $\displaystyle 3{ m }^{ 2 }p-2m{ p }^{ 2 }$
  4. $\displaystyle mp$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \frac { 51{ m }^{ 3 }{ p }^{ 2 }-34{ m }^{ 2 }{ p }^{ 3 } }{ 17mp } $

$\displaystyle =\frac { 17mp\left( 3{ m }^{ 2 }p-2m{ p }^{ 2 } \right)  }{ 17mp } $
$\displaystyle = 3{ m }^{ 2 }p-2m{ p }^{ 2 }$

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 $ax^2+bx+c=0$, then the quadratic equation whose roots are $\cfrac{1}{\alpha}$  and $\cfrac{1}{\beta}$ is

  1. $ax^2+bx+c=0$
  2. $bx^2+ax+c=0$
  3. $cx^2+bx+a=0$
  4. $cx^2+ax+c=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The quadratic equation whose roots are  $\dfrac{1}{\alpha }$ & $\dfrac{1}{\beta }$      is  $x^{2}-\left(\dfrac{1}{\alpha }+\dfrac{1}{\beta }\right)x+\dfrac{1}{\alpha \beta }$
 
=> $ x^{2}-\left(\dfrac{\alpha + \beta }{\alpha \beta}\right)x+\dfrac{1}{\alpha \beta } = 0$
 
also we know that $\alpha +\beta =\dfrac{-b}{a}$ and $\alpha \beta =\dfrac{c}{a}$ as $\alpha,\beta$ are roots of the equation $ax^2+bx+c$
 
=> $ x^{2}-\left(\dfrac{-b }{c}\right)x+\dfrac{a}{c } = 0$
 
=> $ x^{2}+\left(\dfrac{b }{c}\right)x+\dfrac{a}{c } = 0$
 
=>  $  cx^{2}+b x+a = 0 $