Algebra Questions and Answers

Answer: C

One factor is \(p^{2}x - py - z\).

Expand and group the terms so that a common factor of (p + 1) appears:

\[ \begin{aligned} & p^{3}x + p^{2}x - p^{2}y - py - pz - z \\ &= p^{2}x(p + 1) - py(p + 1) - z(p + 1) \\ &= (p + 1)(p^{2}x - py - z) \end{aligned} \]

The other options differ only in signs; multiplying any of them by (p + 1) does not give back the original expression, for example \(p^{2}x + py + z\) would produce +py and +z terms that are not there.

Answer: C

The sum of the first and third integers is 20.

Let the integers be n, n + 1 and n + 2:

\[ \begin{aligned} 3n + 5(n + 1) + 2(n + 2) &= 99 \\ 10n + 9 &= 99 \Rightarrow n = 9 \end{aligned} \]

The integers are 9, 10 and 11 (check: 27 + 50 + 22 = 99), so the first and third add up to 9 + 11 = 20. 18 and 24 are not sums of the first and third of three consecutive integers satisfying the condition.

Answer: C

The sum of the coefficients is 1.

Collect the x² and x terms, remembering that subtracting D changes the sign of each of its terms:

\[ \begin{aligned} x^{2}:\ -2 + 3 - 4 - 1 &= -4 \\ x:\ 12 - 8 + 0 + 1 &= 5 \\ -4 + 5 &= 1 \end{aligned} \]

The usual slip is to change the sign of only the first term of D. Every term of D must change sign, which is what turns its -x into +x and gives -4 and 5; -1 and 3 come from such sign errors.

Answer: D

The solution, x = 8, also satisfies 5(x - 5) = 2(x - 3) + 5.

Solve the given equation:

\[ \begin{aligned} 10x + 60 + 8x - 24 &= 25x - 20 \\ 18x + 36 &= 25x - 20 \\ 56 &= 7x, \quad x = 8 \end{aligned} \]

Now test x = 8 in the options. In 5(x - 5) = 2(x - 3) + 5, the left side is 5(3) = 15 and the right side is 2(5) + 5 = 15, so it holds.

The others fail: 5(5) = 25 but 8 + 5 = 13; 3(19) = 57 but 2(8) + 1 = 17; 2(11) = 22 but 5(3) + 4 = 19.

Answer: C

The expression to subtract is 21y - 24x - 15a.

If A - E = B, then E = A - B. Here A = 5y - 13x - 8a and B = 11x - 16y + 7a:

\[ \begin{aligned} E &= (5y - 13x - 8a) - (11x - 16y + 7a) \\ &= 5y + 16y - 13x - 11x - 8a - 7a \\ &= 21y - 24x - 15a \end{aligned} \]

Check: (5y - 13x - 8a) - (21y - 24x - 15a) = -16y + 11x + 7a, which is B. The option 24x - 21y + a has the signs reversed, as you get by computing B - A, and the other two mishandle the x or a terms.

Answer: D

The value of \(p + q\) is \(-5\).

Put \(t = 2x + 5y\). The left side becomes a quadratic in t, which factorises:

\[ \begin{aligned} t^2 - 5t - 14 &= (t - 7)(t + 2) \\ p &= -7, \quad q = 2 \\ p + q &= -5 \end{aligned} \]

In \((t + p)(t + q)\) the sum p + q is the coefficient of t, here \(-5\). The product pq is \(-14\), which is the distractor, and 9 and 7 come from sign slips.

Answer: C

The expression equals \(-4c\).

Use \(X^2 - Y^2 = (X + Y)(X - Y)\) with \(X = a - b + c\) and \(Y = a - b - c\):

\[ \begin{aligned} X + Y &= 2(a - b), \quad X - Y = 2c \\ X^2 - Y^2 &= 4c(a - b) \\ \frac{4c(a - b)}{b - a} &= -4c \end{aligned} \]

The sign comes from \(b - a = -(a - b)\); dropping it gives 4c, which is not offered, and 2c halves the result.

Answer: A

The product is 3024 \(a^{5}b^{10}\). Read as printed, the 4 and the 14 multiply the two terms, so the coefficients are 6 x 4 for the first and 9 x 14 for the second.

\[ 6 \times 4 \times 9 \times 14 = 3024 \]

The powers of the same letter add when terms are multiplied.

\[ a^{3} \cdot a^{2} = a^{5} \]

\[ b^{8} \cdot b^{2} = b^{10} \]

The other options carry the same letters but coefficients 4524, 3126 and 5044, which no product of 6, 4, 9 and 14 gives. If the 4 and 14 were read as exponents, the answer would be a large power of a and b, not a^5 b^10, so only the multiplier reading fits every option.

Answer: C

The value is \(-\frac{1}{3}\).

\(|9y - 6| = 3\) gives \(9y - 6 = 3\) or \(9y - 6 = -3\), so \(y = 1\) or \(y = \frac{1}{3}\). Evaluate \(y^{2} - \frac{4y}{3}\) for each:

\[ \begin{aligned} y = 1 &: 1 - \tfrac{4}{3} = -\tfrac{1}{3} \\ y = \tfrac{1}{3} &: \tfrac{1}{9} - \tfrac{4}{9} = -\tfrac{1}{3} \end{aligned} \]

Both roots give the same value, so the expression is well defined and equals \(-\frac{1}{3}\). The value \(+\frac{1}{3}\) is a sign slip, 0 would need \(y = \frac{4}{3}\), and 'undefined' would only apply if the two roots gave different values.

Answer: A

The coefficient of \(x^{4}\) is 33.

Combine the factors first: \((x - 1)^{3}(x - 2)^{3} = (x^{2} - 3x + 2)^{3}\). An \(x^{4}\) term arises when the three factors contribute powers adding to 4:

\[ \begin{aligned} x^{2} \cdot x^{2} \cdot 2 &: \ 3 \times 2 = 6 \\ x^{2} \cdot (-3x)(-3x) &: \ 3 \times 9 = 27 \\ \text{total} &= 6 + 27 = 33 \end{aligned} \]

The 3 in each line counts the ways of choosing which factor supplies the odd term. Every term contributing to \(x^{4}\) is positive (the two negative factors of \(-3x\) multiply to a positive), so a negative answer such as −3 is impossible, and 30 and 21 each miss part of the count.