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Algebra 21 Online
OpenStudy (anonymous):

Factor x^3-3x^2-10x+24 Please show work thx

OpenStudy (anonymous):

\[f(x)=x^3-3x^2-10x+24\]\[f(2)=8-12-20+24=0\]so (x-2) is a factor

OpenStudy (anonymous):

\[(x^3-3x^2-10x+24) \div (x-2)=?\] then factorise as you would for a cubic expression.

OpenStudy (anonymous):

Wow thx Dominicng!

OpenStudy (anonymous):

Welcomed.

OpenStudy (anonymous):

x3-3x2-10x+24 Final result : (x + 3) • (x - 2) • (x - 4) Step by step solution : Step 1 : Simplify x3-3x2-10x + 24 Checking for a perfect cube : 1.1 x3-3x2-10x+24 is not a perfect cube Trying to factor by pulling out : 1.2 Factoring: x3-3x2-10x+24 Thoughtfully split the expression at hand into groups, each group having two terms : Group 1: -10x+24 Group 2: x3-3x2 Pull out from each group separately : Group 1: (5x-12) • (-2) Group 2: (x-3) • (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 : 1.3 Find roots (zeroes) of F(x) = x3-3x2-10x+24 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 24. The factor(s) are: of the Leading Coefficient : 1 of the Trailing Constant : 1 ,2 ,3 ,4 ,6 ,8 ,12 ,24 Let us test .... P Q P/Q F(P/Q) Divisor -1 1 -1.00 30.00 -2 1 -2.00 24.00 -3 1 -3.00 0.00 x+3 -4 1 -4.00 -48.00 -6 1 -6.00 -240.00 -8 1 -8.00 -600.00 -12 1 -12.00 -2016.00 -24 1 -24.00 -15288.00 1 1 1.00 12.00 2 1 2.00 0.00 x-2 3 1 3.00 -6.00 4 1 4.00 0.00 x-4 6 1 6.00 72.00 8 1 8.00 264.00 12 1 12.00 1200.00 24 1 24.00 11880.00 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 x3-3x2-10x+24 can be divided by 3 different polynomials,including by x-4 Polynomial Long Division : 1.4 Polynomial Long Division Dividing : x3-3x2-10x+24 ("Dividend") By : x-4 ("Divisor") dividend x3 - 3x2 - 10x + 24 - divisor * x2 x3 - 4x2 remainder x2 - 10x + 24 - divisor * x1 x2 - 4x remainder - 6x + 24 - divisor * -6x0 - 6x + 24 remainder 0 Quotient : x2+x-6 Remainder: 0 Trying to factor by splitting the middle term 1.5 Factoring x2+x-6 The first term is, x2 its coefficient is 1 . The middle term is, +x its coefficient is 1 . The last term, "the constant", is -6 Step-1 : Multiply the coefficient of the first term by the constant 1 • -6 = -6 Step-2 : Find two factors of -6 whose sum equals the coefficient of the middle term, which is 1 . -6 + 1 = -5 -3 + 2 = -1 -2 + 3 = 1 That's it Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -2 and 3 x2 - 2x + 3x - 6 Step-4 : Add up the first 2 terms, pulling out like factors : x • (x-2) Add up the last 2 terms, pulling out common factors : 3 • (x-2) Now add up the four terms of step 3 : (x+3) • (x-2) Which is the desired factorization Final result : (x + 3) • (x - 2) • (x - 4)

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