Squares of Binomials: Additional Practice (Set 1)

Use $a = 7x$ and $b = 2$. Square each term and double their product.

$(7x + 2)^2 = (7x)^2 + 2\cdot(7x)\cdot 2 + 2^2 = 49x^2 + 28x + 4$

The first term is $(7x)^2 = 49x^2$ and the middle term is $28x$.

Use the difference pattern with $a = 6x$ and $b = 1$.

$(6x - 1)^2 = (6x)^2 - 2\cdot(6x)\cdot 1 + 1^2 = 36x^2 - 12x + 1$

The first term is $(6x)^2 = 36x^2$ and the middle term is $-12x$.

The middle term of $(a - b)^2$ is $-2ab$ with $a = 5x$ and $b = 6$.

$-2\cdot(5x)\cdot 6 = -60x$

So the middle term is $-60x$. The value $-30x$ is the single product without doubling.

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

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

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

Use the difference pattern with $a = x$ and $b = 10$.

$(x - 10)^2 = x^2 - 2\cdot x\cdot 10 + 10^2 = x^2 - 20x + 100$

The middle term is $-20x$ and the last term is $+100$.

The area is the side length squared, so compute $(2x + 7)^2$.

$(2x + 7)^2 = (2x)^2 + 2\cdot(2x)\cdot 7 + 7^2 = 4x^2 + 28x + 49$

The middle term is $28x$, and $4x + 14$ is the perimeter, not the area.

Use $a = 3x^2$ and $b = 1$. Squaring $3x^2$ gives $(3x^2)^2 = 9x^4$, and the middle term is $-2\cdot(3x^2)\cdot 1$.

$(3x^2 - 1)^2 = (3x^2)^2 - 2\cdot(3x^2)\cdot 1 + 1^2 = 9x^4 - 6x^2 + 1$

The exponent doubles, so $(3x^2)^2 = 9x^4$.

The outer roots are $\sqrt{36x^2} = 6x$ and $\sqrt{49} = 7$. The middle term has size $2AB$.

$2\cdot 6x\cdot 7 = 84x$

So the trinomial is $36x^2 - 84x + 49 = (6x - 7)^2$.

The middle term of $(x + a)^2$ is $2ax$, and here it is $16x$, so $2a = 16$ gives $a = 8$. Check the last term: $8^2 = 64$, which matches.

$(x + 8)^2 = x^2 + 16x + 64$

So $a = 8$.

The outer roots are $\sqrt{81x^2} = 9x$ and $\sqrt{1} = 1$. The middle should be $2\cdot 9x\cdot 1 = 18x$, which matches.

$81x^2 + 18x + 1 = (9x + 1)^2$

The root of $81x^2$ is $9x$, so $(81x + 1)^2$ is wrong.

Use the difference pattern with $a = 4$ and $b = 3x$.

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

A squared difference keeps a middle term, so it is not the two-term $16 - 9x^2$.

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

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

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