Adding and Subtracting Polynomials: Additional Practice (Set 1)

Track only the $x$ terms. The middle polynomial's $-7x$ flips to $+7x$.

$-7x + 7x + 4x = 4x$

The coefficient of $x$ is $4$.

Add the two known sides, then subtract that from the perimeter.

$(3x - 1) + (4x + 2) = 7x + 1$

$(9x + 4) - (7x + 1) = 2x + 3$

The $x^3$ terms cancel. Combine the rest.

$x^3 - 2x + 1 - x^3 + x^2 - 4 = x^2 - 2x - 3$

Substitute $x = 3$.

$3^2 - 2(3) - 3 = 9 - 6 - 3 = 0$

Three copies of $x^2 - x$ give $3x^2 - 3x$, and two copies of $x^2 + x$ give $2x^2 + 2x$. Subtract.

$3x^2 - 3x - 2x^2 - 2x = x^2 - 5x$

Adding the two equations makes $B$ cancel, since $(A + B) + (A - B) = 2A$.

$2A = 6x^2 + 6x + 2$

Halve each coefficient to get $A$.

$A = 3x^2 + 3x + 1$

Work from the innermost parentheses outward.

$3x - (4x - 5) = -x + 5, \qquad 2x - (-x + 5) = 3x - 5$

Then the outermost subtraction.

$x - (3x - 5) = -2x + 5$

The $x^4$ and $x^3$ terms both cancel when you subtract.

$x^4 + x^3 - x^4 - x^3 + x^2 = x^2$

Only $x^2$ remains, so the degree is $2$.

Adding three copies adds each coefficient three times.

$(x^2 + x^2 + x^2) + (-x - x - x) = 3x^2 - 3x$

Add the first two polynomials; the $x$ terms and the constants cancel.

$(2x^2 - 3x + 1) + (x^2 + 3x - 1) = 3x^2$

Then subtract the third.

$3x^2 - (3x^2 - x) = x$

Collect only the $x^2$ terms, flipping the sign of the last polynomial's $-x^2$.

$-2x^2 + 2x^2 + x^2 = x^2$

The coefficient of $x^2$ is $1$.

Each perimeter is the sum of four equal sides.

$\text{first} = 4x + 12, \qquad \text{second} = 4x - 4$

Subtract; the $x$ terms cancel.

$(4x + 12) - (4x - 4) = 16$

The $x$ terms give $-kx - x = -(k + 1)x$. Set the coefficient equal to $-6$.

$-(k + 1) = -6 \;\Rightarrow\; k + 1 = 6 \;\Rightarrow\; k = 5$