Squares of Binomials: Practice

The outer roots are $\sqrt{9x^2} = 3x$ and $\sqrt{25} = 5$, so the perfect square is $(3x + 5)^2$. Its middle term is $2\cdot 3x\cdot 5$.

$k = 2\cdot 3\cdot 5 = 30$

So $9x^2 + 30x + 25 = (3x + 5)^2$.

Write $52 = 50 + 2$ and apply $(a + b)^2 = a^2 + 2ab + b^2$.

$(50 + 2)^2 = 2500 + 2\cdot 50\cdot 2 + 4 = 2500 + 200 + 4 = 2704$

The middle term $200$ is the part a quick guess would drop.

The outer roots are $\sqrt{36x^2} = 6x$ and $\sqrt{25} = 5$. The middle should be $2\cdot 6x\cdot 5 = 60x$, which matches, with a plus sign.

$36x^2 + 60x + 25 = (6x + 5)^2$

The root of $36x^2$ is $6x$, so $(36x + 5)^2$ is wrong.

Write $99 = 100 - 1$ and apply $(a - b)^2 = a^2 - 2ab + b^2$.

$(100 - 1)^2 = 10000 - 2\cdot 100\cdot 1 + 1 = 10000 - 200 + 1 = 9801$

The subtraction of the middle term $200$ is what a careless guess would miss.

Use $a = 2x^2$ and $b = 3$. Squaring $2x^2$ gives $(2x^2)^2 = 4x^4$, and the middle term is $2\cdot(2x^2)\cdot 3$.

$(2x^2 + 3)^2 = (2x^2)^2 + 2\cdot(2x^2)\cdot 3 + 3^2 = 4x^4 + 12x^2 + 9$

The exponent doubles too, so $(2x^2)^2 = 4x^4$.

For $(x - b)^2$ the middle term is $-2bx$. Matching $-2bx = -12x$ gives $b = 6$, so $c = b^2$.

$c = 6^2 = 36$

Then $x^2 - 12x + 36 = (x - 6)^2$.

Expand the left side and match the middle term. The middle term of $(x + 3)^2$ is $2\cdot x\cdot 3$.

$(x + 3)^2 = x^2 + 6x + 9 \;\Rightarrow\; k = 6$

So $k = 6$, the doubled product of the two terms.

The expression has the shape $a^2 + 2ab + b^2$ with $a = 40$ and $b = 3$, so it is $(40 + 3)^2$.

$(40 + 3)^2 = 43^2 = 1849$

Recognizing the pattern is faster than adding the three pieces separately.

Expand $(x - 7)^2$ with the difference pattern. The middle term is $-2\cdot x\cdot 7 = -14x$ and the last is $+49$.

$(x - 7)^2 = x^2 - 14x + 49$

A squared difference keeps a middle term, so it is not the two-term $x^2 - 49$.

The outer roots are $\sqrt{25x^2} = 5x$ and $\sqrt{36} = 6$. The middle term must be $2AB$.

$2\cdot 5x\cdot 6 = 60x$

So the trinomial is $25x^2 + 60x + 36 = (5x + 6)^2$.

The outer roots are $\sqrt{49x^2} = 7x$ and $\sqrt{16} = 4$, so the square is $(7x - 4)^2$. Its middle term is $-2\cdot 7x\cdot 4$.

$k = 2\cdot 7\cdot 4 = 56$

Then $49x^2 - 56x + 16 = (7x - 4)^2$.

For $9x^2 + 30x + 25$ the outer roots are $3x$ and $5$, and $2\cdot 3x\cdot 5 = 30x$ matches the middle term.

$9x^2 + 30x + 25 = (3x + 5)^2$

The others fail: $20$ is not a perfect square, and $25x$ and $15x$ are not $2\cdot 3x\cdot 5 = 30x$.