Math, asked by ducklings345, 11 months ago

Factorise: x^3 + 5x^2 +7x + 3

Answers

Answered by Anonymous
7

Answer:

Factoring:  x3-5x2+7x-3  

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  7x-3  

Group 2:  x3-5x2  

Pull out from each group separately :

Group 1:   (7x-3) • (1)

Group 2:   (x-5) • (x2)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

2.3    Find roots (zeroes) of :       F(x) = x3-5x2+7x-3

Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  -3.

The factor(s) are:

of the Leading Coefficient :  1

of the Trailing Constant :  1 ,3

Let us test ....

  P    Q    P/Q    F(P/Q)     Divisor

     -1       1          -1.00          -16.00      

     -3      1          -3.00        -96.00      

      1        1           1.00          0.00      x-1  

     3       1            3.00        0.00      x-3  

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that

  x3-5x2+7x-3  

can be divided by 2 different polynomials,including by  x-3

Polynomial Long Division :

2.4    Polynomial Long Division

Dividing :  x3-5x2+7x-3  

                             ("Dividend")

By         :    x-3    ("Divisor")

dividend     x3  -  5x2  +  7x  -  3  

- divisor  * x2     x3  -  3x2          

remainder      -  2x2  +  7x  -  3  

- divisor  * -2x1      -  2x2  +  6x      

remainder             x  -  3  

- divisor  * x0             x  -  3  

remainder                0

Quotient :  x2-2x+1  Remainder:  0  

Trying to factor by splitting the middle term

2.5     Factoring  x2-2x+1  

The first term is,  x2  its coefficient is  1 .

The middle term is,  -2x  its coefficient is  -2 .

The last term, "the constant", is  +1  

Step-1 : Multiply the coefficient of the first term by the constant   1 • 1 = 1  

Step-2 : Find two factors of  1  whose sum equals the coefficient of the middle term, which is   -2 .

     -1    +    -1    =    -2    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -1  and  -1  

                    x2 - 1x - 1x - 1

Step-4 : Add up the first 2 terms, pulling out like factors :

                   x • (x-1)

             Add up the last 2 terms, pulling out common factors :

                    1 • (x-1)

Step-5 : Add up the four terms of step 4 :

                   (x-1)  •  (x-1)

            Which is the desired factorization

Multiplying Exponential Expressions:

2.6    Multiply  (x-1)  by  (x-1)  

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (x-1)  and the exponents are :

         1 , as  (x-1)  is the same number as  (x-1)1  

and   1 , as  (x-1)  is the same number as  (x-1)1  

The product is therefore,  (x-1)(1+1) = (x-1)2  

Final result :

 (x - 1)2 • (x - 3)

Answered by amiratyagi
1

Answer:

Answer:

Here one root is x = 1 so (x - 1) is a factor. i.e

{x}^{3} - {x}^{2} - 3 {x}^{2} + 3x + 2x - 2 =x

3

−x

2

−3x

2

+3x+2x−2=

{x}^{2} (x - 1) - 3x(x - 1) + 2(x - 1) =x

2

(x−1)−3x(x−1)+2(x−1)=

(x - 1)( {x}^{2} - 3x + 2) =(x−1)(x

2

−3x+2)=

(x - 1)( {x}^{2} - 2x - x + 2) =(x−1)(x

2

−2x−x+2)=

(x - 1)(x(x - 2) - 1(x - 2)) =(x−1)(x(x−2)−1(x−2))=

(x - 1)(x - 2)(x - 1) = (x - 1) {}^{2} (x - 2)(x−1)(x−2)(x−1)=(x−1)

2

(x−2)

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