Squares of Binomials: Practice

Apply the pattern with $a = x$ and $b = 5$. Square each term and double their product.

$(x + 5)^2 = x^2 + 2\cdot x\cdot 5 + 5^2 = x^2 + 10x + 25$

The middle term is $10x$, so $x^2 + 25$ is missing it.

Use the difference pattern $(a - b)^2 = a^2 - 2ab + b^2$ with $a = x$ and $b = 4$. Only the middle term is negative.

$(x - 4)^2 = x^2 - 2\cdot x\cdot 4 + 4^2 = x^2 - 8x + 16$

The last term $+16$ stays positive because squaring $-4$ gives $+16$.

Multiply $(a + b)(a + b)$ term by term.

$(a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$

The two equal cross-products combine into $2ab$, so $a^2 + b^2$ is missing the middle term.

The full expansion has three terms, with a middle term of $2ab = 2\cdot x\cdot 4$.

$(x + 4)^2 = x^2 + 8x + 16$

Writing $x^2 + 16$ drops the middle term $8x$, the most common error with this pattern.

The first term is $x^2 = (x)^2$ and the last is $16 = 4^2$, so try $A = x$ and $B = 4$. The middle should be $2AB = 2\cdot x\cdot 4 = 8x$, which matches.

$x^2 + 8x + 16 = (x + 4)^2$

The root of $16$ is $4$, not $8$, so $(x + 8)^2$ is wrong.

The last term is $25 = 5^2$, so $B = 5$, and $A = x$. The middle should be $2\cdot x\cdot 5 = 10x$, and here it is $-10x$, so the sign is negative.

$x^2 - 10x + 25 = (x - 5)^2$

The negative middle term means a difference, so $(x - 5)^2$.

Use $a = x$ and $b = 10$. The middle term is $2\cdot x\cdot 10 = 20x$ and the last is $10^2 = 100$.

$(x + 10)^2 = x^2 + 20x + 100$

The choice $x^2 + 100$ drops the middle term $20x$.

Use the difference pattern with $a = x$ and $b = 1$. The middle term is $-2\cdot x\cdot 1 = -2x$ and the last is $+1$.

$(x - 1)^2 = x^2 - 2x + 1$

The last term is $+1$, since squaring $-1$ gives $+1$.

The middle term of $(a - b)^2$ is $-2ab$. Here $a = x$ and $b = 8$.

$-2\cdot x\cdot 8 = -16x$

So the middle term is $-16x$. The value $-8x$ forgets to double the product.

Multiply $(a - b)(a - b)$ term by term, keeping the signs.

$(a - b)(a - b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2$

The last term is $+b^2$, and only the middle term is negative, so $(a - b)^2$ is not $a^2 - b^2$.

The last term is $1 = 1^2$, so $B = 1$, and $A = x$. Check the middle: $2\cdot x\cdot 1 = 2x$, which matches.

$x^2 + 2x + 1 = (x + 1)^2$

The root of $1$ is $1$, so the factor is $(x + 1)^2$.

For $(x + b)^2$ the middle term is $2bx$. Matching $2bx = 6x$ gives $b = 3$, and the constant is $b^2$.

$b = 3, \qquad b^2 = 3^2 = 9$

So the trinomial is $x^2 + 6x + 9 = (x + 3)^2$.