Math I Expansion and Factoring: Practice Problems and Methods Expanding and factoring underlie almost every calculation in algebra. Learn to pick the right formula quickly and to factor completely.
Basic, Standard, Advanced: Grade 10 · Term 1
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Guide Examples Practice Related
The cross method and substitution When the x 2 x^2 x 2 6 x 2 − 13 x + 6 6x^2 - 13x + 6 6 x 2 − 13 x + 6 cross method : split 6 = 2 × 3 6 = 2 \times 3 6 = 2 × 3 6 = ( − 3 ) × ( − 2 ) 6 = (-3) \times (-2) 6 = ( − 3 ) × ( − 2 ) x x x 2 ⋅ ( − 2 ) + 3 ⋅ ( − 3 ) = − 13 2 \cdot (-2) + 3 \cdot (-3) = -13 2 ⋅ ( − 2 ) + 3 ⋅ ( − 3 ) = − 13
6 x 2 − 13 x + 6 = ( 2 x − 3 ) ( 3 x − 2 ) 6x^2 - 13x + 6 = (2x - 3)(3x - 2) 6 x 2 − 13 x + 6 = ( 2 x − 3 ) ( 3 x − 2 ) When a block of terms repeats, replace it by a single letter . With A = x 2 + 2 x A = x^2 + 2x A = x 2 + 2 x ( x 2 + 2 x ) 2 − 2 ( x 2 + 2 x ) − 3 = ( A − 3 ) ( A + 1 ) (x^2 + 2x)^2 - 2(x^2 + 2x) - 3 = (A - 3)(A + 1) ( x 2 + 2 x ) 2 − 2 ( x 2 + 2 x ) − 3 = ( A − 3 ) ( A + 1 )
Caution
After substituting back, factor as far as possible : ( x 2 + 2 x − 3 ) ( x 2 + 2 x + 1 ) = ( x + 3 ) ( x − 1 ) ( x + 1 ) 2 (x^2 + 2x - 3)(x^2 + 2x + 1) = (x + 3)(x - 1)(x + 1)^2 ( x 2 + 2 x − 3 ) ( x 2 + 2 x + 1 ) = ( x + 3 ) ( x − 1 ) ( x + 1 ) 2
Biquadratics, two letters, symmetric expressions Some biquadratics such as x 4 + x 2 + 1 x^4 + x^2 + 1 x 4 + x 2 + 1 X = x 2 X = x^2 X = x 2 difference of squares :
x 4 + x 2 + 1 = ( x 2 + 1 ) 2 − x 2 = ( x 2 + x + 1 ) ( x 2 − x + 1 ) x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 + x + 1)(x^2 - x + 1) x 4 + x 2 + 1 = ( x 2 + 1 ) 2 − x 2 = ( x 2 + x + 1 ) ( x 2 − x + 1 ) For expressions in two or more letters, arrange in powers of one letter (often the one of lowest degree). For x 2 + x y − 2 y 2 + 2 x + 7 y − 3 x^2 + xy - 2y^2 + 2x + 7y - 3 x 2 + x y − 2 y 2 + 2 x + 7 y − 3 x x x − 2 y 2 + 7 y − 3 = − ( 2 y − 1 ) ( y − 3 ) -2y^2 + 7y - 3 = -(2y - 1)(y - 3) − 2 y 2 + 7 y − 3 = − ( 2 y − 1 ) ( y − 3 )
Symmetric expressions
An expression unchanged by swapping x x x y y y x + y x + y x + y x y xy x y x 2 + y 2 = ( x + y ) 2 − 2 x y x^2 + y^2 = (x + y)^2 - 2xy x 2 + y 2 = ( x + y ) 2 − 2 x y ( x − y ) 2 = ( x + y ) 2 − 4 x y (x - y)^2 = (x + y)^2 - 4xy ( x − y ) 2 = ( x + y ) 2 − 4 x y
Worked examples
Factor completely.
x 2 − 4 x + 3 x^{2} - 4x + 3 x 2 − 4 x + 3 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Find two numbers whose sum is -4 and whose product is 3.
Answer ( x − 3 ) ( x − 1 ) (x - 3)(x - 1) ( x − 3 ) ( x − 1 ) Solution The two numbers with sum − 4 -4 − 4 3 3 3 − 3 -3 − 3 − 1 -1 − 1
Hence
x 2 − 4 x + 3 = ( x − 3 ) ( x − 1 ) \begin{aligned}x^{2} - 4x + 3 &= (x - 3)(x - 1)\end{aligned} x 2 − 4 x + 3 = ( x − 3 ) ( x − 1 )
Factor completely.
4 x 2 − 23 x y + 15 y 2 4x^{2} - 23xy + 15y^{2} 4 x 2 − 23 x y + 15 y 2 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Treat y y y x x x
Answer ( x − 5 y ) ( 4 x − 3 y ) (x - 5y)(4x - 3y) ( x − 5 y ) ( 4 x − 3 y ) Solution Using the cross method with x 2 x^2 x 2 4 4 4 y 2 y^2 y 2 15 15 15
4 x 2 − 23 x y + 15 y 2 = ( x − 5 y ) ( 4 x − 3 y ) \begin{aligned}4x^{2} - 23xy + 15y^{2} &= (x - 5y)(4x - 3y)\end{aligned} 4 x 2 − 23 x y + 15 y 2 = ( x − 5 y ) ( 4 x − 3 y )
Factor completely.
x 4 − 3 x 2 + 9 x^{4} - 3x^{2} + 9 x 4 − 3 x 2 + 9 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Complete the square ( x 2 + 3 ) 2 (x^2 + 3)^2 ( x 2 + 3 ) 2 A 2 − B 2 A^2 - B^2 A 2 − B 2
Answer ( x 2 + 3 x + 3 ) ( x 2 − 3 x + 3 ) (x^{2} + 3x + 3)(x^{2} - 3x + 3) ( x 2 + 3 x + 3 ) ( x 2 − 3 x + 3 ) Solution Rewrite as a difference of squares.
x 4 − 3 x 2 + 9 = ( x 2 + 3 ) 2 − 9 x 2 = ( x 2 + 3 ) 2 − ( 3 x ) 2 x^{4} - 3x^{2} + 9 = (x^2 + 3)^2 - 9x^2 = (x^2 + 3)^2 - (3x)^2 x 4 − 3 x 2 + 9 = ( x 2 + 3 ) 2 − 9 x 2 = ( x 2 + 3 ) 2 − ( 3 x ) 2 Use A 2 − B 2 = ( A + B ) ( A − B ) A^2 - B^2 = (A + B)(A - B) A 2 − B 2 = ( A + B ) ( A − B )
= ( x 2 + 3 x + 3 ) ( x 2 − 3 x + 3 ) = (x^{2} + 3x + 3)(x^{2} - 3x + 3) = ( x 2 + 3 x + 3 ) ( x 2 − 3 x + 3 ) Practice problems
Expand.
( 3 x + 3 y ) ( 3 x − 5 y ) (3x + 3y)(3x - 5y) ( 3 x + 3 y ) ( 3 x − 5 y ) Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Use the distributive law: multiply every term by every term.
Answer 9 x 2 − 6 x y − 15 y 2 9x^{2} - 6xy - 15y^{2} 9 x 2 − 6 x y − 15 y 2 Solution Expand by the distributive law and collect like terms.
( 3 x + 3 y ) ( 3 x − 5 y ) = 9 x 2 − 6 x y − 15 y 2 \begin{aligned}(3x + 3y)(3x - 5y) &= 9x^{2} - 6xy - 15y^{2}\end{aligned} ( 3 x + 3 y ) ( 3 x − 5 y ) = 9 x 2 − 6 x y − 15 y 2
Expand.
( x 2 − 2 x + 5 ) ( x 2 − 2 x + 2 ) (x^{2} - 2x + 5)(x^{2} - 2x + 2) ( x 2 − 2 x + 5 ) ( x 2 − 2 x + 2 ) Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Replace the common part x 2 − 2 x x^{2} - 2x x 2 − 2 x
Answer x 4 − 4 x 3 + 11 x 2 − 14 x + 10 x^{4} - 4x^{3} + 11x^{2} - 14x + 10 x 4 − 4 x 3 + 11 x 2 − 14 x + 10 Solution Let A = x 2 − 2 x A = x^{2} - 2x A = x 2 − 2 x
( A + 5 ) ( A + 2 ) = A 2 + 7 A + 10 (A + 5)(A + 2) = A^2 + 7A + 10 ( A + 5 ) ( A + 2 ) = A 2 + 7 A + 10 Substitute back and expand.
= x 4 − 4 x 3 + 11 x 2 − 14 x + 10 = x^{4} - 4x^{3} + 11x^{2} - 14x + 10 = x 4 − 4 x 3 + 11 x 2 − 14 x + 10
Factor completely.
x 4 − 5 x 2 − 36 x^{4} - 5x^{2} - 36 x 4 − 5 x 2 − 36 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let X = x 2 X = x^2 X = x 2 X X X
Answer ( x + 3 ) ( x − 3 ) ( x 2 + 4 ) (x + 3)(x - 3)(x^{2} + 4) ( x + 3 ) ( x − 3 ) ( x 2 + 4 ) Solution Let X = x 2 X = x^2 X = x 2
X 2 − 5 X − 36 = ( X − 9 ) ( X + 4 ) X^2 - 5X - 36 = (X - 9)(X + 4) X 2 − 5 X − 36 = ( X − 9 ) ( X + 4 ) Substitute back and factor further.
( x 2 − 9 ) ( x 2 + 4 ) = ( x + 3 ) ( x − 3 ) ( x 2 + 4 ) (x^2 - 9)(x^2 + 4) = (x + 3)(x - 3)(x^{2} + 4) ( x 2 − 9 ) ( x 2 + 4 ) = ( x + 3 ) ( x − 3 ) ( x 2 + 4 )