Factoring by grouping is a strategy for factoring polynomials with four or more terms. The main idea is to split the expression into groups, factor out the greatest common factor from each group, and then factor out the common binomial. These problems practice basic grouping, rearranging terms before grouping, and factoring fully using grouping with other factoring patterns.
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 \( \left(4x^2-3\right)(2x-1) \)
\(\,\,\,\,\,\,8x^3-4x^2-6x+3\)
\(\,\,\,\,\,\,4x^2\left(2x-1\right)-3\left(2x-1\right)\)
\(\,\,\,\,\,\,\left(4x^2-3\right)\left(2x-1\right)\)
\(\textbf{3)}\) \(x^3+2x^2-5x-10\)
The answer is \((x+2)\left(x^2-5\right)\)
\(\,\,\,\,\,\,x^3+2x^2-5x-10\)
\(\,\,\,\,\,\,x^2\left(x+2\right)-5\left(x+2\right)\)
\(\,\,\,\,\,\,\left(x^2-5\right)\left(x+2\right)\)
\(\textbf{4)}\) \(3x^3+5x^2+6x+10\)
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 \(\left(3x^2+2\right)\left(x-5\right)\)
\(\,\,\,\,\,\,3x^3-15x^2+2x-10\)
\(\,\,\,\,\,\,3x^2\left(x-5\right)+2\left(x-5\right)\)
\(\,\,\,\,\,\,\left(3x^2+2\right)\left(x-5\right)\)
\(\textbf{8)}\) \(x^3+6x^2+4x+24\)
The answer is \(\left(x^2+4\right)\left(x+6\right)\)
\(\,\,\,\,\,\,x^3+6x^2+4x+24\)
\(\,\,\,\,\,\,x^2\left(x+6\right)+4\left(x+6\right)\)
\(\,\,\,\,\,\,\left(x^2+4\right)\left(x+6\right)\)
\(\textbf{9)}\) \(2x^3+7x^2+6x+21\)
The answer is \(\left(2x+7\right)\left(x^2+3\right)\)
\(\,\,\,\,\,\,2x^3+7x^2+6x+21\)
\(\,\,\,\,\,\,x^2\left(2x+7\right)+3\left(2x+7\right)\)
\(\,\,\,\,\,\,\left(x^2+3\right)\left(2x+7\right)\)
\(\textbf{10)}\) \(5x^3-15x^2+2x-6\)
The answer is \(\left(5x^2+2\right)\left(x-3\right)\)
\(\,\,\,\,\,\,5x^3-15x^2+2x-6\)
\(\,\,\,\,\,\,5x^2\left(x-3\right)+2\left(x-3\right)\)
\(\,\,\,\,\,\,\left(5x^2+2\right)\left(x-3\right)\)
\(\textbf{11)}\) \(6x^3+9x^2+10x+15\)
The answer is \(\left(3x+5\right)\left(2x^2+3\right)\)
\(\,\,\,\,\,\,6x^3+9x^2+10x+15\)
\(\,\,\,\,\,\,3x^2\left(2x+3\right)+5\left(2x+3\right)\)
\(\,\,\,\,\,\,\left(3x^2+5\right)\left(2x+3\right)\)
\(\textbf{12)}\) \(4x^3-8x^2+3x-6\)
The answer is \(\left(4x^2+3\right)\left(x-2\right)\)
\(\,\,\,\,\,\,4x^3-8x^2+3x-6\)
\(\,\,\,\,\,\,4x^2\left(x-2\right)+3\left(x-2\right)\)
\(\,\,\,\,\,\,\left(4x^2+3\right)\left(x-2\right)\)
\(\textbf{13)}\) \(xy+3x+2y+6\)
The answer is \(\left(x+2\right)\left(y+3\right)\)
\(\,\,\,\,\,\,xy+3x+2y+6\)
\(\,\,\,\,\,\,x\left(y+3\right)+2\left(y+3\right)\)
\(\,\,\,\,\,\,\left(x+2\right)\left(y+3\right)\)
\(\textbf{14)}\) \(ab-4a+3b-12\)
The answer is \(\left(a+3\right)\left(b-4\right)\)
\(\,\,\,\,\,\,ab-4a+3b-12\)
\(\,\,\,\,\,\,a\left(b-4\right)+3\left(b-4\right)\)
\(\,\,\,\,\,\,\left(a+3\right)\left(b-4\right)\)
\(\textbf{15)}\) \(12x^3+8x^2-9x-6\)
The answer is \(\left(4x^2-3\right)\left(3x+2\right)\)
\(\,\,\,\,\,\,12x^3+8x^2-9x-6\)
\(\,\,\,\,\,\,4x^2\left(3x+2\right)-3\left(3x+2\right)\)
\(\,\,\,\,\,\,\left(4x^2-3\right)\left(3x+2\right)\)
Challenge Problems
Factor fully.
\(\textbf{16)}\) \( x^5-4x^3+x^2-4 \)
The answer is \( (x+1)(x^2-x+1)(x+2)(x-2) \)
\(\,\,\,\,\,\,x^5-4x^3+x^2-4\)
\(\,\,\,\,\,\,x^3\left(x^2-4\right)+1\left(x^2-4\right)\)
\(\,\,\,\,\,\,\left(x^3+1\right)\left(x^2-4\right)\)
\(\,\,\,\,\,\,\text{Use }a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,\left(x+1\right)\left(x^2-x+1\right)\left(x^2-4\right)\)
\(\,\,\,\,\,\,\text{Use }a^2-b^2=\left(a+b\right)\left(a-b\right)\)
\(\,\,\,\,\,\,\left(x+1\right)\left(x^2-x+1\right)\left(x+2\right)\left(x-2\right)\)
\(\textbf{17)}\) \( 48xy-3xz+80wy-5wz \)
The answer is \( (3x+5w)(16y-z) \)
\(\,\,\,\,\,\,48xy-3xz+80wy-5wz\)
\(\,\,\,\,\,\,3x\left(16y-z\right)+5w\left(16y-z\right)\)
\(\,\,\,\,\,\,\left(3x+5w\right)\left(16y-z\right)\)
\(\textbf{18)}\) \(5x+5x^3+2x^4+2x^6\)
The answer is \( \left(x\right)\left(x^2+1\right)\left(2x^3+5\right)\)
\(\,\,\,\,\,\,5x+5x^3+2x^4+2x^6\)
\(\,\,\,\,\,\,x\left(5+5x^2+2x^3+2x^5\right)\)
\(\,\,\,\,\,\,x\left(5\left(1+x^2\right)+2x^3\left(1+x^2\right)\right)\)
\(\,\,\,\,\,\,x\left(5+2x^3\right)\left(1+x^2\right)\)
\(\,\,\,\,\,\,x\left(x^2+1\right)\left(2x^3+5\right)\)
\(\textbf{19)}\) \(2x^7-8x^5-16x^4+64x^2\)
The answer is \(\,\,\,2x^2\left(x-2\right)^2\left(x+2\right)\left(x^2+2x+4\right)\)
\(\,\,\,\,\,\,2x^7-8x^5-16x^4+64x^2\)
\(\,\,\,\,\,\,2x^2\left(x^5-4x^3-8x^2+32\right)\)
\(\,\,\,\,\,\,2x^2\left(x^3(x^2-4)-8(x^2-4)\right)\)
\(\,\,\,\,\,\,2x^2\left(x^2-4\right)\left(x^3-8\right)\)
\(\,\,\,\,\,\,2x^2\left(x-2\right)\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(\,\,\,\,\,\,2x^2\left(x-2\right)^2\left(x+2\right)\left(x^2+2x+4\right)\)
\(\textbf{20)}\) \(x^4+2x^3-9x^2-18x\)
The answer is \(x\left(x+2\right)\left(x+3\right)\left(x-3\right)\)
\(\,\,\,\,\,\,x^4+2x^3-9x^2-18x\)
\(\,\,\,\,\,\,x\left(x^3+2x^2-9x-18\right)\)
\(\,\,\,\,\,\,x\left(x^2\left(x+2\right)-9\left(x+2\right)\right)\)
\(\,\,\,\,\,\,x\left(x^2-9\right)\left(x+2\right)\)
\(\,\,\,\,\,\,x\left(x+3\right)\left(x-3\right)\left(x+2\right)\)
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