Factor Theorem
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Question 1 of 2
1. Question
Factorise`P(x)=x^3+x^2-14x-24`Hint
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Long Division
Use long division when a polynomial is divided by a binomialFactor Theorem
factor `(P(x))/(x-a)` only if `P(a)=0`Find the first factor using Factor Theorem and trial and error.It is easier to try out numbers that divide to the constant term which is `-24`.`24` can be divided by: `+-` `1,2,3,4,6,8,12`Trying out `a=1`:`P(x)` `=` `x^3+x^2-14x-24` `P(a)` `=` `a^3+a^2-14a-24` Replace `x` with `a` `P(1)` `=` `1^3+1^2-14(1)-24` Substitute `a=1` `=` `1+1-14-24` `=` `-36` Since `P(1)≠0`, `(x-1)` is not a factor of the polynomialTrying out `a=2`:`P(x)` `=` `x^3+x^2-14x-24` `P(a)` `=` `a^3+a^2-14a-24` Replace `x` with `a` `P(2)` `=` `2^3+2^2-14(2)-24` Substitute `a=2` `=` `8+4-28-24` `=` `-40` Since `P(2)≠0`, `(x-2)` is not a factor of the polynomialTrying out `a=-3`:`P(x)` `=` `x^3+x^2-14x-24` `P(a)` `=` `a^3+a^2-14a-24` Replace `x` with `a` `P(-3)` `=` `(-3)^3+(-3)^2-14(-3)-24` Substitute `a=-3` `=` `-27+9+42-24` `=` `0` Since `P(-3)=0`, `(x+3)` is the first factor of the polynomialFind the second factor using Long Division.Substitute components into the formula$$\mathsf{P}$$(Polynomial) `=` `x^3+x^2-14x-24` $$\mathsf{Divisor}$$ `=` `x+3` `=` 
Next, solve for each term of the quotientFirst term of the quotient:Divide the first term of the Polynomial by the first term of the Divisor. Place this above the Polynomial`x^3dividex` `=` `x^2` 
Multiply `x^2` to the divisor. Place this under the Polynomial`x^2``(x+3)` `=` `x^3+3x^2` 
Subtract `x^3+3x^2` and write the difference one line below
Drop down `-14x` and repeat the process to get the second term of the quotient
Second term of the quotient:Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial`-2x^2dividex` `=` `-2x` 
Multiply `-2x` to the divisor. Place this one line below`-2x``(x+3)` `=` `-2x^2-6x` 
Subtract `-2x^2-6x` and write the difference one line below
Drop down `-24` and repeat the process to get the third term of the quotient
Third term of the quotient:Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial`-8xdividex` `=` `-8` 
Multiply `-8` to the divisor. Place this under the Polynomial`-8``(x+3)` `=` `-8x-24` 
Subtract `-8x-24` and write the difference one line below
Since `r=0` and cannot be divided anymore, the quotient is `x^2-2x-8`So far, we have factored the polynomial as `(x+3)``(x^2-2x-8)`Factor out the polynomial further by applying cross method`(x+3)(x^2-2x-8)`
`(x+3)(x+2)(x-4)``(x+3)(x+2)(x-4)` -
Question 2 of 2
2. Question
Factorise`P(x)=2x^3-7x^2-10x+24`Hint
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Correct!
Incorrect
Long Division
Use long division when a polynomial is divided by a binomialFactor Theorem
factor `(P(x))/(x-a)` only if `P(a)=0`Find the first factor using Factor Theorem and trial and error.It is easier to try out numbers that divide to the constant term which is `24`.`24` can be divided by: `+-` `1,2,3,4,6,8,12`Trying out `a=2`:`P(x)` `=` `2x^3-7x^2-10x+24` `P(a)` `=` `2a^3-7a^2-10a+24` Replace `x` with `a` `P(2)` `=` `2(2)^3-7(2)^2-10(2)+24` Substitute `a=2` `=` `2(8)-7(4)-20+24` `=` `16-28-20+24` `=` `16-28-20+24` `=` `-8` Since `P(2)≠0`, `(x-2)` is not a factor of the polynomialTrying out `a=-2`:`P(x)` `=` `2x^3-7x^2-10x+24` `P(a)` `=` `2a^3-7a^2-10a+24` Replace `x` with `a` `P(-2)` `=` `2(-2)^3-7(-2)^2-10(-2)+24` Substitute `a=-2` `=` `2(-8)-7(4)+20+24` `=` `-16-28+20+24` `=` `0` Since `P(-2)=0`, `(x+2)` is the first factor of the polynomialFind the second factor using Long Division.Substitute components into the formula$$\mathsf{P}$$(Polynomial) `=` `2x^3-7x^2-10x+24` $$\mathsf{Divisor}$$ `=` `x+2` `=` 
Next, solve for each term of the quotientFirst term of the quotient:Divide the first term of the Polynomial by the first term of the Divisor. Place this above the Polynomial`2x^3dividex` `=` `2x^2` 
Multiply `2x^2` to the divisor. Place this under the Polynomial`2x^2``(x+2)` `=` `2x^3+4x^2` 
Subtract `2x^3+4x^2` and write the difference one line below
Drop down `-10x` and repeat the process to get the second term of the quotient
Second term of the quotient:Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial`-11x^2dividex` `=` `-11x` 
Multiply `-11x` to the divisor. Place this one line below`-11x``(x+2)` `=` `-11x^2-22x` 
Subtract `-11x^2-22x` and write the difference one line below
Drop down `24` and repeat the process to get the third term of the quotient
Third term of the quotient:Divide the first term of the bottom expression by the first term of the Divisor. Place this above the Polynomial`12xdividex` `=` `12` 
Multiply `12` to the divisor. Place this under the Polynomial`12``(x+2)` `=` `12x+24` 
Subtract `12x+24` and write the difference one line below
Since `r=0` and cannot be divided anymore, the quotient is `2x^2-11x+12`So far, we have factored the polynomial as `(x+2)``(2x^2-11x+12)`Factor out the polynomial further by applying cross method`(x+2)(2x^2-11x+12)`
`(x+2)(2x-3)(x-4)``(x+2)(2x-3)(x-4)`