Arithmetic with Polynomials and Rational Expressions: Proving and Applying Polynomial Identities (CCSS.A-APR.4)
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Common Core High School Algebra › Arithmetic with Polynomials and Rational Expressions: Proving and Applying Polynomial Identities (CCSS.A-APR.4)
Which expansion is equivalent to $(x - y)^4$?
$x^4 - 4x^3y + 4xy^3 - y^4$
$x^4 - 2x^3y + 2xy^3 + y^4$
$x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4$
$(x - y)^2(x^2 + y^2)$
Explanation
Use the binomial theorem with coefficients 1, 4, 6, 4, 1 and alternating signs: $(x - y)^4 = x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4$. Choice B skips the $6x^2y^2$ term and misuses 2 instead of 4 for the outer terms, a common error.
Which identity shows that the triple $(m^2 - n^2, 2mn, m^2 + n^2)$ satisfies the Pythagorean equation $a^2 + b^2 = c^2$?
$(m^2 + n^2)^2 = (m^2 - n^2)^2 + (2mn)^2$
$(m^2 + n^2)^2 = (m^2 - n^2)^2 + (mn)^2$
$(m + n)^2 = (m - n)^2 + (2mn)^2$
$(m^2 + n^2) = (m^2 - n^2)^2 + (2mn)^2$
Explanation
Expand both sides: $(m^2 + n^2)^2 = m^4 + 2m^2n^2 + n^4$ and $(m^2 - n^2)^2 + (2mn)^2 = (m^4 - 2m^2n^2 + n^4) + 4m^2n^2 = m^4 + 2m^2n^2 + n^4$. Choice B uses $(mn)^2$ instead of $(2mn)^2$, dropping the factor 4.
What is the area of the square with side $(3x + 2)$?
$9x^2 + 4$
$9x^2 - 12x + 4$
$9x^2 + 6x + 4$
$9x^2 + 12x + 4$
Explanation
Area is $(3x + 2)^2 = (3x)^2 + 2\cdot(3x)\cdot 2 + 2^2 = 9x^2 + 12x + 4$. Choice A drops the middle term; choice C misuses the middle coefficient (uses 6 instead of 12).
A large square has side $(a + b)$ and a smaller square has side $(a - b)$. What is the difference in their areas?
$a^2 - b^2$
$4ab$
$2ab$
$(a + b)(a - b)$
Explanation
Compute $(a + b)^2 - (a - b)^2 = [(a + b) - (a - b)],[(a + b) + (a - b)] = (2b)(2a) = 4ab$. Choice A confuses the identity with $(a + b)(a - b) = a^2 - b^2$.
Which factorization is equivalent to $x^3 + 8$?
$(x + 2)(x^2 + 2x + 4)$
$(x + 2)^3$
$(x + 2)(x^2 - 2x + 4)$
$(x^2 + 4)(x - 2)$
Explanation
Use $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ with $a = x$, $b = 2$: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$. Choice A has the wrong signs in the quadratic factor (a common mix-up with the difference-of-cubes pattern).
Which expansion is equivalent to $(x - y)^4$?
$x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4$
$x^4 - 4x^3y + 4xy^3 - y^4$
$x^4 - 2x^2y^2 + y^4$
$x^4 - 2x^3y + 2x^2y^2 - 2xy^3 + y^4$
Explanation
By the binomial theorem, $(x - y)^4 = \sum_{k=0}^4 \binom{4}{k} x^{4-k}(-y)^k = x^4 - 4x^3y + 6x^2y^2 - 4xy^3 + y^4$. Choice B skipped the $6x^2y^2$ term and also gave the wrong sign on $y^4$. Choice C treats $(x - y)^4$ like $(x^2 - y^2)^2$, dropping the cubic terms. Choice D misapplies coefficients, using 2 instead of the correct \$4,6,4$.
Which identity shows that $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2$?
$(x^2 + y^2)^2 = x^4 + 4x^2y^2 + y^4$
$(x^2 - y^2)^2 = x^4 + 2x^2y^2 + y^4$
$x^4 + 2x^2y^2 + y^4 = (x^4 - 2x^2y^2 + y^4) + 4x^2y^2$
$(x^2 + y^2)^2 = (x + y)^4$
Explanation
Expand the squares: $(x^2 + y^2)^2 = x^4 + 2x^2y^2 + y^4$ and $(x^2 - y^2)^2 = x^4 - 2x^2y^2 + y^4$. Adding $(2xy)^2 = 4x^2y^2$ gives $x^4 + 2x^2y^2 + y^4 = (x^4 - 2x^2y^2 + y^4) + 4x^2y^2$, which is Choice C. Choice A misuses the middle coefficient (should be 2, not 4). Choice B has the wrong sign in the middle term. Choice D incorrectly treats $x^2 + y^2$ as $x + y$.
What is the area of the square with side $(3x + 2)$?
$9x^2 + 4$
$9x^2 + 6x + 4$
$9x^2 - 12x + 4$
$9x^2 + 12x + 4$
Explanation
Area is $(3x + 2)^2 = (3x)^2 + 2(3x)(2) + 2^2 = 9x^2 + 12x + 4$. Choice A drops the middle $2ab$ term entirely, a common mistake. Choice B misapplies the middle coefficient (should be $2\cdot 3x \cdot 2 = 12x$). Choice C has the wrong sign on the middle term.
For any integer $n$, which identity shows that the product of two consecutive even integers plus 1 is a perfect square?
$n(n+2) = (n+1)^2$
$n(n+2) + 1 = (n+1)^2$
$n(n+2) + 1 = (n+2)^2$
$n(n+2) + 1 = (n^2 + 2n)^2$
Explanation
Compute $n(n+2) + 1 = n^2 + 2n + 1 = (n+1)^2$, so it is a perfect square. Choice A omits the crucial $+1$. Choice C picks the wrong square (off by one). Choice D incorrectly squares the sum $n^2+2n$.
Which factorization is equivalent to $x^4 - y^4$?
$(x - y)(x + y)(x^2 + y^2)$
$(x^2 - y^2)^3$
$(x^2 - y^2)(x^2 - y^2)$
$(x^2 + y^2)(x^2 + y^2)$
Explanation
Use difference of squares twice: $x^4 - y^4 = (x^2)^2 - (y^2)^2 = (x^2 - y^2)(x^2 + y^2)$ and $x^2 - y^2 = (x - y)(x + y)$, giving $(x - y)(x + y)(x^2 + y^2)$. Choice B wrongly treats the expression as a cube. Choice C equals $(x^2 - y^2)^2 = x^4 - 2x^2y^2 + y^4$, not $x^4 - y^4$.