Notes
Practice Problems
Factor the following polynomials.
\(\textbf{1)}\) \( x^3+4x^2+3x+12 \)
The answer is \( (x^2+3)(x+4) \)
\(\,\,\,\,\,\,x^3+4x^2+3x+12\)
\(\,\,\,\,\,\,x^2\left(x+4\right)+3\left(x+4\right)\)
\(\,\,\,\,\,\,\left(x^2+3\right)\left(x+4\right)\)
\(\textbf{2)}\) \( 8x^3-4x^2-6x+3 \)
The answer is \( \left(4x^2-3\right)(2x-1) \)
\(\,\,\,\,\,\,8x^3-4x^2-6x+3\)
\(\,\,\,\,\,\,4x^2\left(2x-1\right)-3\left2x-1\right)\)
\(\,\,\,\,\,\,\left(4x^2-3\right)\left(2x-1\right)\)
\(\textbf{3)}\) \(x^3+2x^2-5x-10\)
The answer is \((x+2)\left(x^2-5\right)\)
\(\,\,\,\,\,\,x^3+2x^2-5x-10\)
\(\,\,\,\,\,\,x^2\left(x+2\right)-5\left(x+2\right)\)
\(\,\,\,\,\,\,\left(x^2-5\right)\left(x+2\right)\)
![Link to Youtube Video Solving Question Number 3](https://andymath.com/wp-content/uploads/2020/08/play-video.jpg")
\(\textbf{4)}\) \(3x^3+5x^2+6x+10\)
The answer is \(\left(3x+5\right)\left(x^2+2\right)\)
\(\,\,\,\,\,\,3x^3+5x^2+6x+10\)
\(\,\,\,\,\,\,x^2(3x+5)+2(3x+5)\)
\(\,\,\,\,\,\,\left(3x+5\right)\left(x^2+2\right)\)
\(\textbf{5)}\) \(3x^3+24x^2+2x+16\)
The answer is \(\left(3x^2+2\right)\left(x+8\right)\)
\(\,\,\,\,\,\,3x^3+24x^2+2x+16\)
\(\,\,\,\,\,\,3x^2\left(x+8\right)+2\left(x+8\right)\)
\(\,\,\,\,\,\,\left(3x^2+2\right)\left(x+8\right)\)
\(\textbf{6)}\) \(11x-22-5x^3+10x^2\)
The answer is \(\left(x-2\right)\left(-5x^2+11\right)\)
\(\,\,\,\,\,\,11x-22-5x^3+10x^2\)
\(\,\,\,\,\,\,11\left(x-2\right)-5x^2\left(x-2\right)\)
\(\,\,\,\,\,\,\left(11-5x^2\right)\left(x-2\right)\)
\(\textbf{7)}\) \(3x^3-15x^2+2x-10\)
The answer is \(\left(3x^2+2\right)\left(x-5\right)\)
\(\,\,\,\,\,\,3x^3-15x^2+2x-10\)
\(\,\,\,\,\,\,3x^2\left(x-5\right)+2\left(x-5\right)\)
\(\,\,\,\,\,\,\left(3x^2+2\right)\left(x-5\right)\)
Challenge Problems
Factor fully.
\(\textbf{8)}\) \( x^5-4x^3+x^2-4 \)
The answer is \( (x+1)(x^2-x+1)(x+2)(x-2) \)
\(\,\,\,\,\,\,x^5-4x^3+x^2-4\)
\(\,\,\,\,\,\,x^3\left(x^2-4\right)+1\left(x^2-4\right)\)
\(\,\,\,\,\,\,\left(x^3+1\right)\left(x^2-4\right)\)
\(\text{Hint: } a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,\left(x+1\right)\left(x^2-x+1\right)\left(x^2-4\right)\)
\(\text{Hint: } a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(x+1\right)\left(x^2-x+1\right)\left(x+2\right)\left(x-2\right)\)
\(\textbf{9)}\) \( 48xy-3xz+80wy-5wz \)
The answer is \( (3x+5w)(16y-z) \)
\(\,\,\,\,\,\,48xy-3xz+80wy-5wz\)
\(\,\,\,\,\,\,3x\left(16y-z\right)+5w\left(16y-z\right)\)
\(\,\,\,\,\,\,\left(3x+5w\right)\left(16y-z\right)\)
\(\textbf{10)}\) \(5x+5x^3+2x^4+2x^6\)
The answer is \( \left(x\right)\left(x^2+1\right)\left(2x^3+5\right)\)
\(\,\,\,\,\,\,5x+5x^3+2x^4+2x^6\)
\(\,\,\,\,\,\,\left(x\right)\left(5+5x^2+2x^3+2x^5\right)\)
\(\,\,\,\,\,\,\left(x\right)\left(5\left(1+x^2\right)+2x^3\left(1+x^2\right)\right)\)
\(\,\,\,\,\,\,\left(x\right)\left(5+2x^3\right)\left(1+x^2\right)\)
\(\textbf{11)}\) \(2x^7-8x^5-16x^4+64x^2\)
The answer is \(\,\,\,2x^2\left(x-2\right)^2\left(x+2\right)\left(x^2+2x+4\right)\)
\(\,\,\,\,\,\,2x^7-8x^5-16x^4+64x^2\)
\(\,\,\,\,\,\,2x^2\left(x^5-4x^3-8x^2+32\right)\)
\(\,\,\,\,\,\,2x^2\left(x^3(x^2-4)-8(x^2-4)\right)\)
\(\,\,\,\,\,\,2x^2\left(x^2-4\right)\left(x^3-8\right)\)
\(\,\,\,\,\,\,2x^2\left(x-2\right)\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(\,\,\,\,\,\,2x^2\left(x-2\right)^2\left(x+2\right)\left(x^2+2x+4\right)\)
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