Ontario · MPM2D · Grade 10

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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Practice Questions (13)

Question 1 · Factoring Area Applications

A rectangular garden has an area represented by the expression $$A = x^2 + 11x + 28$$ square metres. Determine expressions for the length and width of the garden. If $$x = 5$$ m, find the actual dimensions and the area of the garden.
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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 $$(2x + 5)$$ cm and $$(3x + 4)$$ cm. A small rectangular cutout of dimensions $$(x + 1)$$ cm by $$(x + 2)$$ cm is removed. Find a simplified expression for the area of the remaining metal plate.
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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 $$A = 4x^2 + 20x + 25$$ square centimetres. If the frame is square, find an expression for the side length. If $$x = 8$$ cm, find the perimeter of the outer frame.
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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: $$6x^2 + 15x - 9$$
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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?
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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?
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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?
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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: $$15a^3b^2 - 25a^2b^3 + 10a^2b^2$$
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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: $$x(y - 4) - (y - 4)$$
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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): $$-2x^2 + 8x$$
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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: $$x^2 + 100$$
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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: $$x^2 - 12x + 36$$
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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: $$25x^2 - 70x + 49$$
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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

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