Factorise x3 - 23x2 + 142x - 120

x3 - x2 - 22x+ 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
divide  x3-23x2+142x-120
• -12
divide  x3-23x2+142x-120 by( x-1) by long dividtion

• -14
THIS IS THE ANSWER TO FIRST QUESTION OF ANURAG FROM EAST POINT SCHOOL.
P(x)=x^3- 23x^2+ 142x- 120
g(x)=x-1
when we have finished  the division theorem we get,
r(x)=0,  q(x)= x^2- 22x+ 120.
• -3
x323x​2+142x-120
let p(x)=​ x323x​2+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- 23x​2+142x-120
​
by long division

x3- 23x​2+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
x3-23x2+142x-120=
x3+142=143
-23x2-120=-143
=143-143=0
• -2
if x=  1/√(3-√2) find √x + 1/√x
• -3
math

• 3
geo
• -1
use photomath
• 7