Unit 3 · Ontario MPM2D

Ontario Grade 10 Polynomials & Factoring Practice Questions

Expanding in reverse. Pull out the GCF first, then spot the pattern: simple trinomial (a=1), complex (a≠1), difference of squares, or perfect square. Area questions show why it matters.

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What you need to know

GCF first — always check before anything else.
x²+bx+c — two numbers that multiply to c and add to b.
ax²+bx+c (a≠1) — numbers that multiply to a·c and add to b, then split and group.
Special cases — a²−b²=(a−b)(a+b); a²±2ab+b²=(a±b)².

Practice questions with solutions

13 questions. Try each one on paper, then open the worked solution.

Question 1 · Factoring Area Applications

A rectangular garden has an area represented by the expression A=x2+11x+28A = x^2 + 11x + 28A=x2+11x+28 square metres. Determine expressions for the length and width of the garden. If x=5x = 5x=5 m, find the actual dimensions and the area of the garden.

Show worked solution

Let the length and width of the garden be l and w.
Formula/model:
A = x² + 11x + 28
Solve:
Factor the trinomial: A = (x + 7)(x + 4), so the dimensions are (x + 7) and (x + 4).
Substitute x = 5: length = 12 m, width = 9 m, and area = 12 × 9 = 108.
Therefore, dimensions=(x+4) by (x+7), length=12 m, width=9 m, area=108 m^2

Answer: dimensions=(x+4) by (x+7), length=12 m, width=9 m, area=108 m^2

Question 2 · Factoring Area Applications

A rectangular metal plate has outer dimensions of (2x+5)(2x + 5)(2x+5) cm and (3x+4)(3x + 4)(3x+4) cm. A small rectangular cutout of dimensions (x+1)(x + 1)(x+1) cm by (x+2)(x + 2)(x+2) cm is removed. Find a simplified expression for the area of the remaining metal plate.

Show worked solution

Let A_outer be the area of the outer rectangle and A_inner be the area of the cutout.
Formula/model:
A_shaded = (2x + 5)(3x + 4) - (x + 1)(x + 2)
Solve:
Expand the areas: A_outer = 6x² + 23x + 20 and A_inner = x² + 3x + 2.
Subtract to find the remaining area: A_shaded = (6x² + 23x + 20) - (x² + 3x + 2) = 5x² + 20x + 18.
Therefore, 5x^2+20x+18

Answer: 5x^2+20x+18

Question 3 · Factoring Area Applications

A picture frame has an outer area represented by A=4x2+20x+25A = 4x^2 + 20x + 25A=4x2+20x+25 square centimetres. If the frame is square, find an expression for the side length. If x=8x = 8x=8 cm, find the perimeter of the outer frame.

Show worked solution

Let s be the side length of the square frame, and P be the perimeter.
Formula/model:
s = √(4x² + 20x + 25)
P = 4s
Solve:
Factor the area perfect square trinomial: 4x² + 20x + 25 = (2x + 5)², so side length s = 2x + 5.
Substitute x = 8: s = 2(8) + 5 = 21.
Calculate the perimeter: P = 4 × 21 = 84.
Therefore, side=2x+5, perimeter=84 cm

Answer: side=2x+5, perimeter=84 cm

Question 4 · Complex Trinomials (a ≠ 1)

Factor fully: 6x2+15x−96x^2 + 15x - 96x2+15x−9

Show worked solution

Always check for a GCF first. The GCF of 6x², 15x, and -9 is 3.
Factor out 3:
3(2x² + 5x - 3)
Now, decompose the trinomial inside. Product = 2 × (-3) = -6, sum = 5.
Two numbers: 6 and -1.
3(2x² + 6x - x - 3)
= 3[2x(x + 3) - 1(x + 3)]
= 3(2x - 1)(x + 3).

Answer: 3(2x-1)(x+3)

Question 5 · Factoring Concepts

Which of the following binomials is a difference of squares that can be factored over the integers?

Show worked solution

A difference of squares must be in the form a² - b² where both terms are perfect squares separated by subtraction.
4x² - 25 = (2x)² - 5² is a difference of squares.
x² + 9 and 9x² + 16 are sums of squares (not factorable over integers).
x² - 7 is not factorable over integers because 7 is not a perfect square. The correct answer is c.

Answer: $4x^2 - 25$

Question 6 · Factoring Concepts

Which of the following trinomials is a perfect square trinomial?

Show worked solution

A perfect square trinomial is of the form a² + 2ab + b² = (a+b)².
For x² + 6x + 9: a = x, b = 3. The middle term is 2ab = 2(x)(3) = 6x, and the last term is b² = 3² = 9.
Thus, x² + 6x + 9 = (x + 3)², which is a perfect square trinomial. The correct answer is a.

Answer: $x^2 + 6x + 9$

Question 7 · Factoring Concepts

Which of the following trinomials cannot be factored over the integers?

Show worked solution

To factor x² + 2x + 5, we need two integers that multiply to 5 and add to 2. The only integer factors of 5 are 1 and 5 (sum = 6) or -1 and -5 (sum = -6). No integers add up to 2. Thus, it is not factorable over integers. The correct answer is c.

Answer: $x^2 + 2x + 5$

Question 8 · Common Factoring (GCF)

Factor the greatest common factor (GCF) out of the trinomial: 15a3b2−25a2b3+10a2b215a^3b^2 - 25a^2b^3 + 10a^2b^215a3b2−25a2b3+10a2b2

Show worked solution

Identify the GCF of the three terms:
For coefficients 15, -25, and 10, the GCF is 5.
For variable a (a³, a², a²), the lowest power is a².
For variable b (b², b³, b²), the lowest power is b².
Thus, GCF = 5a²b².
Divide each term by 5a²b²:
15a³b² / 5a²b² = 3a
-25a²b³ / 5a²b² = -5b
10a²b² / 5a²b² = 2
Write as a product: 5a²b²(3a - 5b + 2).

Answer: 5a^2b^2(3a-5b+2)

Question 9 · Common Factoring (GCF)

Factor out the common binomial factor: x(y−4)−(y−4)x(y - 4) - (y - 4)x(y−4)−(y−4)

Show worked solution

In the expression x(y - 4) - (y - 4), there is an implicit coefficient of 1 in front of the second term: x(y - 4) - 1(y - 4).
The common binomial factor is (y - 4).
Factor out (y - 4):
Dividing the first term by (y - 4) leaves x.
Dividing the second term by (y - 4) leaves -1.
Write as a product: (y - 4)(x - 1).

Answer: (y-4)(x-1)

Question 10 · Common Factoring (GCF)

Factor fully by extracting a negative greatest common factor (GCF): −2x2+8x-2x^2 + 8x−2x2+8x

Show worked solution

Identify the negative GCF of the terms -2x² and 8x:
Since the leading coefficient is negative, we factor out a negative coefficient GCF.
The GCF of -2 and 8 is -2 (taking the negative).
The GCF of x² and x is x.
Thus, GCF = -2x.
Divide each term by -2x:
-2x² / -2x = x
8x / -2x = -4
Write as a product: -2x(x - 4).

Answer: -2x(x-4)

Question 11 · Difference of Squares & Perfect Squares

Factor fully: x2+100x^2 + 100x2+100

Show worked solution

The binomial x² + 100 is a sum of squares. A sum of squares a² + b² cannot be factored over the integers (it requires imaginary numbers, which are not used in grade 10 math). Therefore, it is not factorable over integers.

Answer: not factorable over integers

Question 12 · Difference of Squares & Perfect Squares

Factor the perfect square trinomial: x2−12x+36x^2 - 12x + 36x2−12x+36

Show worked solution

Check for a perfect square trinomial: a² - 2ab + b² = (a - b)².
The first term is x² → a = x.
The last term is 36 = 6² → b = 6.
Since the middle term is negative (-12x), use the subtraction identity.
Verify: -2(x)(6) = -12x, which matches.
Factored form: (x - 6)².

Answer: (x-6)^2

Question 13 · Difference of Squares & Perfect Squares

Factor fully: 25x2−70x+4925x^2 - 70x + 4925x2−70x+49

Show worked solution

Check for a perfect square trinomial: a² - 2ab + b² = (a - b)².
First term: 25x² = (5x)² → a = 5x.
Last term: 49 = 7² → b = 7.
Middle term is negative: -2ab = -2(5x)(7) = -70x, which matches.
Factored form: (5x - 7)².

Answer: (5x-7)^2

Common mistakes

Missing the GCF and trying to factor what's left.
Sign errors when c is negative.
Trying to factor a²+b² over integers — it doesn't factor.
Stopping after partial factoring; check again.

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Ontario Grade 10 Polynomials & Factoring Practice Questions

What you need to know

Practice questions with solutions

Common mistakes

Polynomials & Factoring Unit Test

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