9th Class Mathematics Polynomials Standard Formula

Standard Formula

Category : 9th Class

*         Standard Formula

 

  • \[{{\text{(a}+\text{b})}^{\text{2}}}={{\text{a}}^{\text{2}}}+\text{2ab}+{{\text{b}}^{\text{2}}}\]
  • \[~{{(a-b)}^{\text{2}}}={{\text{a}}^{\text{2}}}-\text{2ab}+{{\text{b}}^{\text{2}}}\]
  • \[{{\text{a}}^{\text{2}}}-{{\text{b}}^{\text{2}}}=(\text{a}+\text{b})(\text{a}-\text{b})\]
  • \[{{\text{(a}+\text{b})}^{\text{2}}}+{{(\text{a}-\text{b})}^{\text{2}}}=\text{2(}{{\text{a}}^{\text{2}}}+{{\text{b}}^{\text{2}}}\text{)}\]
  • \[{{\text{(a}+\text{b)}}^{\text{2}}}-{{(\text{a}-\text{b})}^{\text{2}}}=\text{4ab}\]
  • \[{{\text{(a}+\text{b}+\text{c)}}^{\text{2}}}-{{\text{a}}^{\text{2}}}+{{\text{b}}^{\text{2}}}+{{\text{c}}^{\text{2}}}+\text{2ab}+\text{2bc}+\text{2ca}\]
  • \[{{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)\]
  • \[{{\text{(a}-\text{b})}^{\text{3}}}={{\text{a}}^{\text{3}}}+{{\text{b}}^{\text{3}}}-\text{3ab}(\text{a}-\text{b)}\]
  • \[{{\text{a}}^{3}}-{{\text{b}}^{\text{3}}}=(\text{a}+\text{b)}-({{a}^{2}}-ab+{{b}^{2}})\]
  • \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\]
  • \[{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=(a-b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\]
  • \[{{a}^{3}}-{{b}^{3}}-{{c}^{3}}-3abc=\frac{1}{2}(a+b+c)[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}]\]
  • \[(y+a)(y+b)={{y}^{2}}+(a+b)y+ab\]
  • \[(y+a)(y+b)(y+c)={{y}^{3}}+(a+b+c){{y}^{2}}\] \[+(ab+bc+ca)y+abc\]      

 

 

 

 

  • Polynomials are used to make polynomial rings which is a central concept in abstract algebra and algebraic geometry in advanced mathematics.
  • Degree and coefficient of a polynomials decide true nature of the graph. If the degree n of a polynomial is even then the arms of the graph are either both Up or both down. If the degree n is odd then one arm of the graph is up and one is down. If the coefficient \[{{a}_{n}}\] is positive then the right-arm of the graph is up. If the coefficient \[{{a}_{n}}\] is negative then the right arm of the graph is down.      

 

 

 

  • A polynomia \[p(x)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+....+{{a}_{n}}{{x}^{n}}\], where \[{{a}_{0}}+{{a}_{1}}+{{a}_{2}}....{{a}_{n}}\] are constant with \[{{a}_{n}}\ne 0\] and n is a non-negative integer.
  • The highest power of the variable is called degree of the polynomial.
  • We can classify polynomials by two methods

  (a) According to degree of polynomials as linear polynomial quadratic polynomial cubic polynomial biquadratic polynomial etc.

  (b) According to number of terms as monomials binomials trinomials etc.

  • \[(x-a)\] is said to be factor of \[p(x)\] if and only if \[p(a)=0\]    

 

 

 

  If \[(3-2x)\]and\[(5x+8)\] are factors of \[(-10{{x}^{2}}+hx-k)\] then the value of h and k are____.

(a) (-31, 24)                                        

(b) (31, 24)

(c) (-31, -24)                                      

(d) (31, -24)

(e) None of these  

 

Answer: (b)

Explanation:

Since \[(3-2x)\] is a factor of \[-10{{x}^{2}}+hx-k\]. Therefore, for \[x=\frac{3}{2}\] the value of given polynomial is zero.

i.e. \[10{{x}^{2}}+hx-k=0\]

or \[-10{{\left( \frac{3}{2} \right)}^{2}}+h\times \frac{3}{2}-k=0\]

or, \[3h-2k=45\]                          ..................... (i)

Similarly, \[(5x+8)\] is a factor of \[-10{{x}^{2}}+hx-k\]

That implies \[\text{8h}+\text{5k}=\text{128}\]               ............. (ii)

Solve equation (i) and (ii) and get the result.  

 

 

  If a and b are two positive integers and \[a+b+ab+1=77\] then find the possible value of a + b.

(a) 16                                                    

(b) 18

(c) 22                                                    

(d) 23

(e) None of these  

 

Answer: (a)

Explanation:

\[\text{a}+\text{b}+\text{ab}+\text{l}=\text{77}\] \[\Rightarrow \]\[\text{(1}+\text{a})(\text{1}+\text{b})=\text{11}\times \text{7}\] \[\Rightarrow \]If (1 + a) = 11 then (1 + b) = 7 and vice versa.  

 

 

  Which one of the following statements is false for the method of factorization of an algebraic expression?

(a) By using standard identities

(b) By using remainder theorem

(c) Taking out common factor from two or more than two terms

(d) Taking out common factor from a group of terms

(e) None of these

 

Answer: (b)  

 

 

  If \[\text{a}+\text{b}+\text{c}=\text{1},\text{ ab}+\text{bc}+\text{ca}=-\text{1},\text{ abc}=-\text{1}\], then \[{{\text{a}}^{\text{3}}}\text{+}{{\text{b}}^{\text{3}}}\text{+}{{\text{c}}^{\text{3}}}\] is equal to;

(a) -2                                    

(b) 2

(c) -1                                     

(d) 1                      

(e) None of these

 

Answer: (d)

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