Chapter Closure
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Exercise 100 Page 252
a Use an area model and a diamond problem to factor the expression.
b Use an area model and a diamond problem to factor the expression.
c This can be rewritten in the form a^2+2ab+b^2.
d This is a difference of squares.
e This is written in the form a^2+2ab+b^2.
f This is a difference of squares.
g Which expression is written, or can be rewritten, in the form a^2-b^2 or a^2+2ab+b^2?
a (x+4)(x+2)
b 3(x+1)(x+4)
c 2(x+1)^2
d (x+6)(x-6)
e (x+a)^2
f (x+y)(x-y)
g C: Perfect square trinomial
D: Difference of squares
E: Perfect square trinomial
F: Difference of squares
a Let's use a generic rectangle and a diamond problem to solve this. We know that x^2 and 8 goes into the lower left and upper right corners of the generic rectangle.
Notice that the product and sum are both positive. This means both factors must be positive. |c|c|c|r|c| [-1em] Product & ax(bx) & ax+bx & Sum & 6x? [0.2em] [-1em] 8x^2 & x(8x) & x+8x& 9x & * [0.1em] 8x^2 & 2x(4x) & 2x+4x& 6x & ✓ When one term is 2x and the other is 4x, we have a product of 8x^2 and a sum of 6x. Now we can complete the diamond and generic rectangle.
To factor the quadratic expression we add each side of the area model and multiply the sums. x^2+6x+8=(x+4)(x+2)
b Like in Part A, we can use an generic rectangle and a diamond problem to factor the expression. However, to make our lives a little bit easier we can start by factoring out 3.
To fill in the remaining two rectangles, we need to find two x-terms with a sum of 5x and a product of 4x^2.
Notice that the product and the sum are both positive. This means both factors must be positive. |c|c|c|c|c| [-1em] Product & ax(bx) & ax+bx & Sum & 5x? [0.2em] [-1em] 4x^2 & 2x(2x) & 2x+2x& 4x & * [0.1em] 4x^2 & 1x(4x) & 1x+4x& 5x & ✓ When one term is x and the other is 4x, we have a product of 4x^2 and a sum of 5x. Now we can complete the diamond and area model.
To factor the quadratic expression we add each side of the area model and multiply the sums. 3x^2+15x+12=3(x+1)(x+4)
c Like in Part B, we will start by factoring out 2.
2(x^2+2x+1)
WritePow
Write as a power
2(x^2+2x+1^2)
SplitIntoFactors
Split into factors
2(x^2+2(x)(1)+1^2)
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
2(x+1)^2
d Examining the expression, we see that x^2 and 36 are both perfect squares. Since 36 is subtracted from x^2 this is a difference of squares, which means we can factor it.
e Like in Part D, we notice that the trinomial is written in the form a^2+2ab+b^2, which means it can be factored as (a+b)^2.
f Examining the expression we see that x^2 and y^2 are both squares. Since y^2 is subtracted from x^2, this is a difference of squares which means we can factor it.
g Whenever an expression is written in the form a^2-b^2 or a^2±2ab+b^2, it can be rewritten in the following ways.
a^2-b^2=(a+b)(a-b) a^2+2ab+b^2=(a+b)^2 a^2-2ab+b^2=(a-b)^2 In Parts C through F, we have either a perfect square trinomial (C and E) or a difference of squares (D and F).
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