Tag: algebraic formulae - expansion of squares

Questions Related to algebraic formulae - expansion of squares

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

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

  1. $5, 4$

  2. $4,9$

  3. $4,5$

  4. $1, -1$

Show answer
Correct Option: C
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
$a=4$   ;$b=5$

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

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

  1. True

  2. False

Show answer
Correct Option: B
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.

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

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

  1. $\displaystyle { m }^{ 2 }p$

  2. $\displaystyle m{ p }^{ 2 }$

  3. $\displaystyle 3{ m }^{ 2 }p-2m{ p }^{ 2 }$

  4. $\displaystyle mp$

Show answer
Correct Option: C
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 } $
$\displaystyle = 3{ m }^{ 2 }p-2m{ p }^{ 2 }$

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

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

  1. $ax^2+bx+c=0$

  2. $bx^2+ax+c=0$

  3. $cx^2+bx+a=0$

  4. $cx^2+ax+c=0$

Show answer
Correct Option: C
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 $