Solving Polynomial equations by Factoring

Solving Polynomial equations by Factor sums and differences of cubes

by Ayah Alshaiksalem

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Multiple Choice

Find the product $$\left(x-3\right)\ \left(x^2+\ 3x+9\right)$$  

• $$x^{3\ }+\ 3x^2\ +9\$$  

• $$x^3-3\ x^2-9x\ +9$$  

• $$x^{3\ }+3x^2+9x\ -3x^2-9x\ -\ 9\ =x^{3\ }-\ 27$$   $$=\ x^{3\ ^{\ }}-\ 27\$$  

• $$x+3$$  

Multiple Choice

SAT Question

If y = $$x^3+2x+5\ and\ z=x^2+7x+1,\ what\ is\ 2y+z\ in\ the\ terms\ of\ x?\$$  

• $$3x^3+11x\ +\ 11\$$  

• $$2x^{3\ }+x^2+9x+6$$  

• $$2x^3+x^2+11x+11$$  

• $$2x^3+\ 2x^2+18x\ +12$$  

​Instructed Practice 1 : Objective: solve a polynomial equation by factor the difference of cubs.

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Difference of two cubes

Multiple Choice

Difference of two cubes $$a^{3\ }-\ b^{3\ }=\ \left(a-b\right)\left(a^2+\ ab+b^2\right)\longrightarrow\ \ \ \ \ \ \ \left(x\right)^{3\ }-\ \left(2\right)^3\ =\ \left(x-2\right)\ \left(\ x^2+--\ +\ --\right)$$  

• $$2x+\ \left(2\right)^2$$  

• 4x+ 16

• 2x - 8

• 2x - $$\left(2\right)^2$$  

Multiple Choice

Difference of two cubes $$a^{3\ }-\ b^{3\ }=\ \left(a-b\right)\left(a^2+\ ab+b^2\right)\ \ \ \ \ \ \ \ \ \ x^{3\ }-\ \left(5\right)^3\ =\ \left(x-5\right)\ \left(\ -\ +\ -\ +-\right)$$  

• $$x^2+5x+\ \left(5\right)^2$$  

• $$x^2\ -5x+\ \left(5\right)^2$$  

• $$\left(5\right)^2\ -5x-\ x^2$$  

• 2x - $$\left(2\right)^2$$  

Open Ended

How can you check that an expression has been factored correctly ??

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​How can you check that an expression has been factored correctly ??

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Multiply the factors to check that the product is the same as the original expression (write it down in your notebook)

Instructed Practice 2 : Objective: solve a polynomial equation by factor the sum of cubs.

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Difference of two cubes

​The sum of two cubes

Multiple Choice

The Sum of two cubes $$x^3+\ y^{3\ }=\ \left(x+y\right)\left(x^{2\ }-\ xy\ +\ y^2\right)$$  

$$\ \ \ \ \ x^3\ +\ \left(3\right)^3$$ =  

• $$\left(x-3\right)\left(x^2\ +\ 3x+9\right)$$  

• $$\left(x+3\right)\left(x^2-3x\ +\ 9\right)$$  

Multiple Choice

The Sum of two cubes $$x^3+\ y^{3\ }=\ \left(x+y\right)\left(x^{2\ }-\ xy\ +\ y^2\right)$$  

$$\ \ \ \ \ x^3\ +\ \left(1\right)^3$$ =

• $$x^3+\ 1^{3\ }=\ \left(x-1\right)\left(x^{2\ }-\ x\ -\ 1\right)$$  

• $$x^3+1^3=\ \left(x+1\right)\left(x^2-x+1\right)$$  

• $$x^3+\ 1^3\ =\left(x-1\right)\left(x^2+x\ +1\right)$$  

Multiple Choice

Factor completely $$3\ x^3\ +\ 3\ \left(6\right)^3$$  

• $$3\left(x+6\right)\left(x^2-6x+36\right)$$  

• $$3\left(x+6\right)\left(x^2\ +\ 6x+36\right)$$  

• $$3\left(x-\ 6\right)\left(x^2\ +\ 6x-36\right)$$  

Poll

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