Special Factorizations: Chapter Test

Solution

Correct answer: A

You chose B: Uses a single $ab = 8x$ for the middle term instead of doubling it to $2ab$.

You chose C: Drops the middle term entirely, treating $(a + b)^2$ as $a^2 + b^2$.

You chose D: Copies the last term as $b = 8$ instead of squaring it to $b^2 = 64$.

Solution

Correct answer: C

You chose A: Reads a difference of squares as a squared binomial, giving two same-sign factors instead of opposite signs.

You chose B: Writes the roots in the reversed order, so the product is $121 - x^2$, the negative of the given expression.

You chose D: Mistakes the difference for a sum of squares and stops, missing that $a^2 - b^2$ always factors.

Solution

Correct answer: D

You chose A: Multiplies only the denominator by $\sqrt{3}$, leaving the numerator as $6$ over $3$.

You chose B: Takes $\sqrt{3}\cdot\sqrt{3}$ as $9$ instead of $3$, so the denominator is too large.

You chose C: Multiplies the numerator by $\sqrt{3}$ but cancels the denominator to $1$ rather than to $3$.

Solution

Correct answer: B

You chose A: Keeps the trinomial's middle term positive, when SOAP makes its sign the opposite of the binomial's.

You chose C: Doubles the middle term to $2ab = 4x$, importing the perfect-square pattern instead of the cube trinomial's single $ab$.

You chose D: Reads $x^3 + 8$ as $(x + 2)^3$, treating a sum of cubes as a cubed sum.

Solution

Correct answer: C

You chose A: Pairs the constants $5$ and $1$ with the wrong leading terms, as $(2x + 5)(x + 1)$, producing a middle term of $7x$.

You chose B: Makes the same wrong pairing with negative constants, as $(2x - 5)(x - 1)$, producing a middle term of $-7x$.

You chose D: Uses the AC pair $10$ and $1$ directly as the factor constants, as $(x + 10)(x + 1)$, which changes the constant term to $10$.

Solution

Correct answer: A

You chose B: Checks only that the outer terms are perfect squares, missing that the middle term must equal $2AB = 24x$, not $12x$.

You chose C: Accepts the middle term without checking that the last term $12$ is a perfect square.

You chose D: Allows a negative last term, when a perfect-square trinomial's last term is always positive.

Solution

Correct answer: D

You chose A: Stops after pulling out the $2$, leaving the difference of squares $x^2 - 25$ unfactored.

You chose B: Factors $x^2 - 25$ as a perfect square instead of a difference of squares.

You chose C: Drops the common factor of $2$ instead of carrying it into the factorization.

Solution

Correct answer: B

You chose A: Computes the conjugate product as $a^2 + b = 28$ instead of $a^2 - b = 22$.

You chose C: Multiplies the numerator by the original denominator $5 - \sqrt{3}$ rather than by the conjugate $5 + \sqrt{3}$.

You chose D: Squares $\sqrt{3}$ as $9$ instead of $3$, giving a denominator of $16$.

Solution

Correct answer: C

You chose A: Makes the trinomial's middle term negative, when SOAP makes its sign the opposite of the binomial's minus.

You chose B: Doubles the middle term to $2ab = 24x$ instead of the cube trinomial's single $ab = 12x$.

You chose D: Fails to square the whole term $3x$, writing $3x^2$ instead of $(3x)^2 = 9x^2$.

Solution

Correct answer: A

You chose B: Swaps the constants $5$ and $3$ between the two factors, rebuilding a different polynomial.

You chose C: Stops at the grouped line without factoring out the shared binomial $(x + 5)$.

You chose D: Mismatches the sign of the shared binomial, factoring the first pair as $x^2(x - 5)$.

Solution

Correct answer: B

You chose A: Pulls out the $3$ but leaves the perfect-square trinomial $x^2 + 10x + 25$ unfactored.

You chose C: Drops the common factor of $3$ instead of carrying it into the factorization.

You chose D: Factors the trinomial as a difference of squares, ignoring its middle term $30x$.

Solution

Correct answer: D

You chose A: Factors the difference as a squared binomial, giving same-sign factors instead of opposite signs.

You chose B: Uses the coefficients $36$ and $25$ themselves instead of their square roots $6$ and $5$.

You chose C: Takes $\sqrt{25y^2}$ as $5$ instead of $5y$, dropping the variable.

Solution

Correct answer: A

You chose B: Factors the difference of squares but leaves each cube factor unfactored.

You chose C: Factors each cube as $(x \pm 2)(x^2 + 4)$, using $b^2$ alone for the trinomial and dropping its middle term.

You chose D: Reads $x^6 - 64$ as a perfect square $(x^3 - 8)^2$ rather than a difference of two squares.

Solution

Correct answer: C

You chose A: Treats the conjugate product as $1$ instead of $7 - 2 = 5$.

You chose B: Computes the denominator as $7 + 2 = 9$ instead of $7 - 2 = 5$.

You chose D: Multiplies the numerator by a copy of the denominator instead of by the conjugate $\sqrt{7} - \sqrt{2}$.

Solution

Correct answer: B

You chose A: Mismatches the sign of the shared binomial, factoring the first pair as $3x^2(2x + 3)$.

You chose C: Factors $+2$ out of the second pair instead of $-2$, so the last two terms come back with the wrong signs.

You chose D: Stops at the grouped line without pulling out the shared binomial $(2x - 3)$.

Solution

Correct answer: D

You chose A: Pulls out the $2$ but leaves the sum of cubes $x^3 + 8$ unfactored.

You chose B: Keeps the trinomial's middle term positive, when SOAP makes its sign the opposite of the binomial's plus.

You chose C: Drops the common factor of $2$ instead of carrying it through the factorization.

Solution

Correct answer: C

You chose A: Uses a plus sign, when the negative middle term $-30xy$ needs $(3x - 5y)$.

You chose B: Ignores the middle term and factors the outer terms as a difference of squares.

You chose D: Takes $\sqrt{25y^2}$ as $5$ instead of $5y$, dropping the variable.

Solution

Correct answer: A

You chose B: Computes the denominator as $5 + 2 = 7$ instead of the difference of squares $5 - 2 = 3$.

You chose C: Squares the numerator as $5 + 2 = 7$, dropping the middle term $2\sqrt{10}$.

You chose D: After the denominator becomes $3$, squares the denominator expression $\sqrt{5} - \sqrt{2}$ instead of the numerator, putting $7 - 2\sqrt{10}$ on top.

Solution

Correct answer: D

You chose A: Reverses the signs of the split pair, giving a middle term of $+17x$.

You chose B: Swaps the constants $5$ and $1$ between the two factors, as $(3x - 1)(4x + 5)$, giving a middle term of $+11x$.

You chose C: Splits the leading coefficient as $6$ and $2$ without checking the middle term against $ac$.

Solution

Correct answer: B

You chose A: Uses a single $ab = 5\cdot 7 = 35$ instead of the doubled middle term $2ab$.

You chose C: Adds the roots $5 + 7 = 12$ instead of doubling their product.

You chose D: Doubles the product a second time, giving $140$ instead of $70$.