Squares of Binomials: Practice

Use the difference pattern with $a = 3x$ and $b = 2$. Square each term and subtract twice their product.

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

The first term is $(3x)^2 = 9x^2$ and the middle term is $-12x$.

Use $a = 4x$ and $b = 1$. Square each term and double their product.

$(4x + 1)^2 = (4x)^2 + 2\cdot(4x)\cdot 1 + 1^2 = 16x^2 + 8x + 1$

The first term is $(4x)^2 = 16x^2$ and the middle term is $8x$.

The pattern works with two variables. Take $a$-term $= 2a$ and $b$-term $= 5b$.

$(2a + 5b)^2 = (2a)^2 + 2\cdot(2a)\cdot(5b) + (5b)^2 = 4a^2 + 20ab + 25b^2$

Each outer term is squared in full, and the middle term $20ab$ is twice their product.

The first term is $9x^2 = (3x)^2$ and the last is $4 = 2^2$, so $A = 3x$ and $B = 2$. Check the middle: $2\cdot 3x\cdot 2 = 12x$, which matches.

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

The root of $9x^2$ is $3x$, not $9x$, so $(9x + 2)^2$ is wrong.

The outer roots are $\sqrt{25x^2} = 5x$ and $\sqrt{4} = 2$. The middle should be $2\cdot 5x\cdot 2 = 20x$, and here it is $-20x$, so the sign is negative.

$25x^2 - 20x + 4 = (5x - 2)^2$

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

The middle term is $2ab$ with $a = 4x$ and $b = 3$.

$2\cdot(4x)\cdot 3 = 24x$

So the middle term is $24x$. The value $12x$ is the single product $4x\cdot 3$ without doubling.

Use $a = 3x$ and $b = 4$. Square each term and double their product.

$(3x + 4)^2 = (3x)^2 + 2\cdot(3x)\cdot 4 + 4^2 = 9x^2 + 24x + 16$

The middle term is $24x$, not $12x$, because the product $3x\cdot 4$ is doubled.

The last term of $(a - b)^2$ is $b^2$. Here $b = 5$.

$(-5)^2 = 25$

Squaring $-5$ gives $+25$, so the constant term is positive.

A perfect square would be $(3x + 5)^2 = 9x^2 + 30x + 25$, which has a middle term. The expression $9x^2 + 25$ has only two terms.

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

Without the middle term $30x$ it is not a perfect-square trinomial. (Note $25$ is a perfect square, so that reason is false.)

Use $(a - b)^2$ with $a = 6$ and $b = x$.

$(6 - x)^2 = 6^2 - 2\cdot 6\cdot x + x^2 = 36 - 12x + x^2$

The last term $x^2$ is positive, since squaring $-x$ gives $+x^2$.

The outer roots are $\sqrt{49x^2} = 7x$ and $\sqrt{1} = 1$. The middle should be $2\cdot 7x\cdot 1 = 14x$, which matches.

$49x^2 + 14x + 1 = (7x + 1)^2$

The root of $49x^2$ is $7x$, so $(49x + 1)^2$ is wrong.

The last term is $49 = 7^2$, so $B = 7$, and $A = x$. The middle term must be $2AB$.

$2\cdot x\cdot 7 = 14x$

So the trinomial is $x^2 + 14x + 49 = (x + 7)^2$.