a Examining the expression we see that 4, x^2, and 49 are all perfect squares. Since 49 is subtracted from 4x^2 this must be a difference of squares, which means we can factor it.
b Let's use a generic rectangle and a diamond problem to solve this. We know that 2x^2 and 12 goes into the lower left and upper right corners of the generic rectangle.
To fill in the remaining two rectangles we need two x-terms that have a sum of 11x and a product of 24x^2.
Notice that the product and the sum are both positive. This means both factors must be positive.
|c|c|c|r|c|
[-1em]
Product & ax(bx) & ax+bx & Sum & 11x? [0.2em]
[-1em]
24x^2 & x(24x) & x+24x& 25x & * [0.1em]
24x^2 & 2(12x) & 2x+12x& 14x & * [0.1em]
24x^2 & 3x(8x) & 3x+8x& 11x & âś“ [0.1em]
When one term is 3x and the other is 8x, we have a product of 24x^2 and a sum of 11x. Now we can complete the diamond and generic rectangle.
To factor the quadratic expression we add each side of the generic rectangle and multiply the sums.
2x^2+11x+12=(2x+3)(x+4)
c Like in Part A, both 4x^2 and 25 are perfect squares.
4x^2+25 ⇔ (2x)^2+5^2
However, since 25 is not subtracted from 4x^2 we cannot factor this like we could when we had a difference of squares in Part A.
d Examining the expression, we notice that the trinomial can be rewritten in the form a^2+2ab+b^2, which means it can be factored as (a+b)^2.
e The expression consists of four terms which means we need a 2* 2 area model. Let's create such a model and include the first term in the lower left rectangle, which can only be factored as x* y.
Next, we will factor -4y in the upper left rectangle and 2x in the lower right rectangle. They belong in these spots because of where we placed x and y. Note that this leaves the upper right rectangle left for -8.
Finally, let's calculate the products and label the length and width of the area model.
Now we can write the factored expression.
(x-4)(y+2)
f Like in Part A, we can rewrite this as a difference of squares and thereby factor it.
