x3 - x2 - 22x2 + 22x + 120x + 120
x2 (x-1) - 22x (x-1) + 120 (x-1)
(x-1) (x2 - 22x + 120)
(x-1) (x2 -12x - 10x +120)
(x-1) [ x(x-12) - 10(x-12)]
(x-1) (x-12) (x-10) Ans.
- 0
x 3 -23x 2 +142x-120
Final result :
(x - 1) • (x - 10) • (x - 12)
Step by step solution :
Step 1 :
Raise x to the 2nd power
Exponentiation :
Equation at the end of step 1 :
(((x 3 ) - (23 • x 2)) + 142x) - 120
Step 2 :
Raise x to the 3rd power
Exponentiation :
Equation at the end of step 2 :
((x 3 - 23x 2 ) + 142x) - 120
Step 3 :
Simplify x 3-23x 2 +142x - 120
Checking for a perfect cube :
3.1 x 3 -23x 2 +142x-120 is not a perfect cube
Trying to factor by pulling out :
3.2 Factoring: x 3 -23x 2 +142x-120
Thoughtfully split the expression at hand into groups, each group
having two terms :
Group 1: 142x-120
Group 2: x 3 -23x 2
Pull out from each group separately :
Group 1: (71x-60) • (2)
Group 2: (x-23) • (x 2 )
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 :
3.3 Find roots (zeroes) of : F(x) = x3 -23x 2 +142x-120
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 -120.
The factor(s) are:
of the Leading Coefficient : 1
of the Trailing Constant : 1 ,2 ,3 ,4 ,5 ,6 ,8 ,10 ,12 ,15 , etc
Let us test ....
P Q P/Q F(P/Q) Divisor
-1 1 -1.00 -286.00
-2 1 -2.00 -504.00
-3 1 -3.00 -780.00
-4 1 -4.00 -1120.00
-5 1 -5.00 -1530.00
-6 1 -6.00 -2016.00
-8 1 -8.00 -3240.00
-10 1 -10.00 -4840.00
-12 1 -12.00 -6864.00
-15 1 -15.00 -10800.00
1 1 1.00 0.00 x-1
2 1 2.00 80.00
3 1 3.00 126.00
4 1 4.00 144.00
5 1 5.00 140.00
6 1 6.00 120.00
8 1 8.00 56.00
10 1 10.00 0.00 x-10
12 1 12.00 0.00 x-12
15 1 15.00 210.00
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
x 3 -23x 2 +142x-120
can be divided by 3 different polynomials,including by x-12
Polynomial Long Division :
3.4 Polynomial Long Division
Dividing : x 3 -23x 2 +142x-120
("Dividend")
By : x-12 ("Divisor")
dividend x 3 - 23x 2 + 142x - 120
- divisor * x 2 x 3 - 12x 2
remainder - 11x 2 + 142x - 120
- divisor * -11x 1 - 11x 2 + 132x
remainder 10x - 120
- divisor * 10x 0 10x - 120
remainder 0
Quotient : x 2 -11x+10 Remainder: 0
Trying to factor by splitting the middle term
3.5 Factoring x 2 -11x+10
The first term is, x2 its coefficient is 1 .
The middle term is, -11x its coefficient is -11 .
The last term, "the constant", is +10
Step-1 : Multiply the coefficient of the first term by the constant
1 • 10 = 10
Step-2 : Find two factors of 10 whose sum equals the coefficient
of the middle term, which is -11 .
-10 + -1 = -11 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the
two factors found in step 2 above, -10 and -1
x2 - 10x - 1x - 10
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-10)
Add up the last 2 terms, pulling out common factors :
1 • (x-10)
Step-5 : Add up the four terms of step 4 :
(x-1) • (x-10)
Which is the desired factorization
Final result :
(x - 1) • (x - 10) • (x - 12)
Processing ends successfully
#srikar
- -17
x3- 23x2+142x-120
let p(x)= x3- 23x2+142x-120
g(x)=x-1
0=x-1
x=1
p(1)=(1)3-23(1)2+142(1)-120
=1-23+142-120
=143-143
=0
Hence x-1 is a factor of x3- 23x2+142x-120
by long division
x3- 23x2+142x-120 = (x-1)(x2+22x+120)
=(x-1)(x2+10x+12x+120)
=(x-1)(x(x+10) 12(x+10))
=(x-1)(x+12)(x+10)
let p(x)= x3- 23x2+142x-120
g(x)=x-1
0=x-1
x=1
p(1)=(1)3-23(1)2+142(1)-120
=1-23+142-120
=143-143
=0
Hence x-1 is a factor of x3- 23x2+142x-120
by long division
x3- 23x2+142x-120 = (x-1)(x2+22x+120)
=(x-1)(x2+10x+12x+120)
=(x-1)(x(x+10) 12(x+10))
=(x-1)(x+12)(x+10)
- 28