Factorising and Solving Quadratics Practice Quiz (20 questions)

1. A student changes $5x+10$ into $5(x+2)$. Which operation has the student performed?

A. Expanding by multiplying factors out

B. Rearranging terms without changing their form

C. Factorising by taking out a common factor

D. Combining like terms into one term

Answer: (C)

Explanation

Factorising reverses expansion by writing an expression as a product of factors, here taking out 5. Expanding multiplies factors out, while combining like terms and rearranging do not create this product form.

2. A student writes $12y^2+18y=3y(4y+6)$. What change is needed to make the factorisation complete?

A. Take out an additional factor of y

B. Take out another common factor of 2

C. Rearrange the two terms inside the bracket

D. Expand the bracket to check the expression

Answer: (B)

Explanation

A factorisation is complete when every common factor has been taken outside the brackets; 4y and 6 still share a factor of 2. Expanding reverses factoring, y is not common to both bracket terms, and rearranging does not remove a common factor.

3. To factorise $6px+9py+4qx+6qy$ by grouping, which sequence should be used?

A. Pair terms with common factors, factor each pair, then take out the common bracket

B. Pair terms without checking for common factors, then expand each pair

C. Factor each pair and leave the matching bracket in both terms

D. Take out one common factor from all terms, then multiply the remaining terms

Answer: (A)

Explanation

Grouping works by pairing terms that share a factor, factoring each pair, and then taking out the resulting common bracket. A single overall factor or expanding does not perform this sequence, and leaving the matching bracket in both terms is incomplete.

4. A monic quadratic is $x^2+9x+20$. Which pair of constants gives its factorisation using the product-and-sum method?

A. 1 and 20

B. 2 and 10

C. 4 and 5

D. 3 and 6

Answer: (C)

Explanation

For a monic quadratic, the constants must multiply to 20 and add to 9; 4 and 5 meet both conditions, giving $(x+4)(x+5)$. The other pairs either have the wrong sum or the wrong product.

5. When factoring $4x^2+11x+6$ by splitting the middle term and grouping, which two numbers should be used to split $11x$?

A. 1 and 10

B. 2 and 3

C. 3 and 8

D. 4 and 6

Answer: (C)

Explanation

The method seeks numbers with product $ac=4\times6=24$ and sum $b=11$, so the split is $3x+8x$. The pair 2 and 3 multiplies to the constant 6, while 4 and 6 and 1 and 10 fail the required sum.

6. A student needs to factor $9p^2-16q^2$. Which factorisation uses the difference-of-squares identity correctly?

A. $(9p-16q)(p+q)$

B. $(3p+4q)(3p+4q)$

C. $(3p-4q)(3p+4q)$

D. $(3p-4q)^2$

Answer: (C)

Explanation

A difference of squares factors as the difference of the square roots multiplied by their sum, giving $(3p-4q)(3p+4q)$. Squaring one bracket or using two same-sign brackets does not produce the original difference.

7. To factorise $6x^3+9x^2-15x$, which first step and follow-up method should be used?

A. Group the terms as a difference of two squares

B. Take out $3x$, then factorise the remaining quadratic

C. Take out $3$, then factorise the remaining cubic

D. Take out $x$, then stop with a quadratic inside the bracket

Answer: (B)

Explanation

A cubic of the form $ax^3+bx^2+cx$ is handled by taking out the common factor of $x$ (or a multiple of it) and then factorising the remaining quadratic. Here, taking out $3x$ leaves a quadratic to factorise; leaving it unfactored misses the second step.

8. How many real solutions can a quadratic equation have?

A. One or two

B. Exactly one

C. Exactly two

D. Zero, one, or two

Answer: (D)

Explanation

A quadratic equation can have zero, one, or two solutions. The claim of exactly one reflects the comparison with linear equations, while excluding zero or one omits valid possibilities for quadratics.

9. A factored equation is $(y-6)(2y+1)=0$. Which step correctly uses the zero-product rule to find its solutions?

A. Set $y-6=2y+1$

B. Set $y-6=1$ and $2y+1=1$

C. Set $y-6=0$

D. Set $y-6=0$ and $2y+1=0$

Answer: (D)

Explanation

When a product equals zero, at least one factor must equal zero, so each factor is set to zero to find the possible solutions. Equating the factors or setting a factor to one does not apply the zero-product rule.

10. A student is solving $z^2+z-20=7z$ by factoring. What sequence of steps should the student use?

A. Rearrange to zero, factorise, and stop

B. Rearrange to zero, factorise, set each factor to zero, and solve

C. Rearrange to zero, set each term to zero, and solve

D. Factorise as written, set each factor to zero, and solve

Answer: (B)

Explanation

To solve a factorisable quadratic, first rearrange it so one side is zero, then factorise, set each factor equal to zero, and solve the resulting equations. Factoring before rearranging or setting individual terms to zero does not follow this procedure.

11. A student solves $3x^2-5x-2=0$ with the quadratic formula. Which pair of solutions should the student obtain?

A. $x=\frac{-5\pm\sqrt{49}}{6}$

B. $x=\frac{1}{3},\;x=-2$

C. $x=2,\;x=-\frac{1}{3}$

D. $x=\frac{5\pm\sqrt{49}}{3}$

Answer: (C)

Explanation

For $a=3$, $b=-5$, and $c=-2$, the formula gives $x=\frac{5\pm\sqrt{25+24}}{6}=\frac{5\pm7}{6}$, hence $2$ and $-\frac{1}{3}$. The other responses reflect changing the denominator, losing one sign, or using the wrong sign for $b$.

12. A student uses the quadratic formula to find exact values of $x$ and then uses them to calculate a dependent quantity. To preserve accuracy, what should the student do?

A. Use the positive root in later calculations and omit the other root

B. Round the discriminant first and use the approximate square root

C. Round each root first and use the rounded roots in later calculations

D. Keep exact roots during the calculation and round the final values

Answer: (D)

Explanation

Intermediate values should remain exact because rounding them can affect dependent results; rounding is done at the end. Rounding roots or the discriminant early introduces approximation, while dropping a root ignores one of the equation's solutions.

13. To solve $x^2+8x-1=0$ by completing the square, which expression correctly replaces $x^2+8x$?

A. $(x+4)^2+16$

B. $(x+8)^2-64$

C. $(x+4)^2-16$

D. $(x+4)^2-8$

Answer: (C)

Explanation

For a monic quadratic, $x^2+bx$ becomes $(x+\frac{b}{2})^2-(\frac{b}{2})^2$; with $b=8$, this is $(x+4)^2-16$. Using $8$ instead of its half, adding rather than subtracting the correction, or subtracting $8$ fails to preserve the original expression.

14. A function is written as $f(x)=(x-5)^2-7$. What is its minimum value, and at which input does it occur?

A. Minimum value $-7$ at $x=-5$

B. Minimum value $-7$ at $x=5$

C. Minimum value $7$ at $x=5$

D. Minimum value $5$ at $x=-7$

Answer: (B)

Explanation

The squared term is nonnegative and reaches zero at $x=5$, so the minimum of the whole expression is its offset, $-7$. The square cannot lower the value below the offset; the input is $5$, not $-5$.

15. A rectangular garden has a length 5 meters greater than its width and an area of 84 square meters. After defining the width as a variable and forming a quadratic, which approach correctly identifies the garden’s dimensions?

A. Solve the quadratic and discard any root that gives an impossible dimension

B. Solve the quadratic and accept every root as a possible width

C. Solve the quadratic and use the same root for length and width

D. Solve a linear equation by adding 5 to the area

Answer: (A)

Explanation

A word problem is modeled by defining a variable, translating the relationships into an equation, solving the resulting quadratic, and checking its roots against the context. Adding 5 to the area bypasses the model, using the same value for both dimensions ignores their difference, and accepting every root skips the contextual check.

16. A calculation for a physical setup gives two mathematical roots for a length, $-2$ cm and $6$ cm. Which value can represent the length in the setup?

A. $6$ cm

B. Both $-2$ cm and $6$ cm

C. Neither value, because there are two roots

D. $-2$ cm

Answer: (A)

Explanation

A valid physical length cannot be negative, so the root $-2$ cm must be rejected and $6$ cm retained. Accepting both roots confuses mathematical solutions with contextual ones, while choosing by size or rejecting both does not apply the length constraint.

17. A student must solve the simultaneous equations $y=3x-2$ and $y=x^2+1$. Which method is appropriate for this linear–nonlinear pair?

A. Substitute the linear expression for $y$ into the nonlinear equation

B. Graph both equations instead of forming an algebraic equation

C. Eliminate variables using the method for two linear equations

D. Solve the linear equation alone for its value of $x$

Answer: (A)

Explanation

For a pair with one linear and one nonlinear equation, substitution lets the linear expression be inserted into the nonlinear equation. Elimination is associated with two linear equations, solving one equation alone does not satisfy both, and graphing is not the stated method.

18. For the system $y=2x+1$ and $y=x^2-3$, what sequence should be used to find all ordered-pair solutions?

A. Substitute the nonlinear expression into the linear equation, then stop at the x-values

B. Substitute $2x+1$ into the nonlinear equation, solve the quadratic, then find each paired $y$

C. Eliminate $y$, solve a linear equation, then calculate a single paired value

D. Solve the nonlinear equation for $y$, then use one root as both coordinates

Answer: (B)

Explanation

The procedure is to substitute the linear expression into the nonlinear equation, solve the resulting quadratic for the possible x-values, and substitute each x-value back to find its y-value. Finding y first, using elimination for this pair, or stopping at x-values leaves the system unsolved.

19. Which description best characterizes Srinivasa Ramanujan's mathematical background and contributions?

A. A formally trained Indian mathematician who specialized in image compression and solved problems considered unsolvable

B. A formally trained Indian mathematician who contributed to geometry and algebra but avoided unsolved problems

C. A self-taught Belgian mathematician who contributed to analysis and number theory and solved routine problems

D. A self-taught Indian mathematician who contributed to analysis and number theory and solved problems considered unsolvable

Answer: (D)

Explanation

Ramanujan was self-taught and Indian, contributed to analysis and number theory, and solved problems then considered unsolvable. The formal-training distractors misstate his background, the image-compression distractor confuses his work with Daubechies's, and the routine-problem distractor weakens the stated significance of his work.

20. A researcher identifies a Belgian physicist and mathematician whose best-known work concerns image compression technology. Who is the researcher describing?

A. G. H. Hardy

B. Srinivasa Ramanujan

C. Ingrid Daubechies

D. A mathematician known for analyzing paintings

Answer: (C)

Explanation

Ingrid Daubechies is identified as a Belgian physicist and mathematician best known for work on image compression technology. Ramanujan is associated with analysis and number theory, Hardy is identified as his collaborator, and art authentication is a related application rather than the stated defining work.