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Quadratic trinomials and factorisation

This test focuses on transforming and factorising quadratic trinomials. Tasks cover expansion, completing the square, vertex form and selecting an appropriate algebraic method for a given expression.

Questions: 10Estimated time: 20 minutes

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Question 1out of 10

When factoring using roots x₁ and x₂, why is the form (x − x₁)(x − x₂) used?

The roots must be multiplied by 2

The constant term is ignored

The signs in the factors are opposite to the root signs

The factors use the same signs as the roots

Question 2out of 10

Expand (x − (-1))(x − (2)).

x² + (-1)x + (2)

x² + (-2)x + (-1)

x² + (1)x + (-2)

x² + (-1)x + (-2)

To expand the brackets, multiply each term in the first pair of brackets by each term in the second:

(x − a)(x − b) = x² − bx − ax + ab.

Combine the like terms:

(x − a)(x − b) = x² − (a + b)x + ab.

Multiply x · x.
Combine the two terms containing x: −bx − ax.
Multiply the constant terms: (−a)(−b) = ab.

Pay close attention to the signs. If a negative number appears in brackets, for example x − (−3), the expression is equal to x + 3.

Question 3out of 10

What does a positive discriminant of a quadratic trinomial mean?

It has exactly one root

It cannot be a quadratic trinomial

It has no real roots

It has two real roots and can be factored into two linear factors

Question 4out of 10

For y = (x − (1))² + (3), give the vertex.

(1, 3)

(1, 4)

(1, -3)

(2, 3)

(-1, 3)

(3, 1)

The vertex form of a parabola is y = a(x − m)² + n.

Its vertex coordinates can be read directly:

V(m, n).

Pay close attention to the sign inside the brackets. For example, if the expression is x − (−3), the x-coordinate of the vertex is −3, not 3.

Question 5out of 10

Write x² + 8x + 16 as a perfect square.

(x + 4)²

(x + 8)²

x² + 16

(x − 4)²

Use the perfect-square identity:

(x + a)² = x² + 2ax + a².

Check that the final term is the square of a and that the coefficient of x equals 2a. When both conditions hold, the trinomial can be written as (x + a)².

You can also find a by dividing the coefficient of x by 2.

Question 6out of 10

Factorise x² + 13x + 42.

(x − 6)(x − 7)

(x + 13)(x + 42)

(x − 6)(x + 7)

(x + 6)(x + 7)

Question 7out of 10

Factorise x² − 8x + 15.

(x − 3)(x − 5)

(x + 3)(x + 5)

(x + 3)(x − 5)

(x − 8)(x − 15)

The quadratic trinomial x² − sx + p can be factorised into two factors:

(x − a)(x − b).

Find two numbers a and b that satisfy both conditions:

their sum equals s: a + b = s;
their product equals p: a · b = p.

First list the factor pairs of p, then select the pair whose sum equals s.

You can check the result by expanding the brackets: (x − a)(x − b) = x² − (a + b)x + ab.

Question 8out of 10

Factorise x² + 11x + 24.

(x − 8)(x − 3)

(x + 11)(x + 24)

(x + 8)(x + 3)

(x − 8)(x + 3)

Question 9out of 10

Find k so that x² + kx + 100 equals (x + 10)².

Use the perfect-square identity:

(x + a)² = x² + 2ax + a².

Compare it with the given trinomial x² + kx + c:

the constant term c must equal a²;
the middle term kx must match 2ax.

Therefore, to find k, multiply a by 2. Substitute the result for k to check that the two expressions are identical.

Question 10out of 10

Write x² + 8x + 16 as a perfect square.

(x + 8)²

(x + 4)²

x² + 16

(x − 4)²

Use the perfect-square identity:

(x + a)² = x² + 2ax + a².

Check that the final term is the square of a and that the coefficient of x equals 2a. When both conditions hold, the trinomial can be written as (x + a)².

You can also find a by dividing the coefficient of x by 2.

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Sample questions

Factorise x² − ___x + ___.
Expand (x − (___))(x − (___)).
Find k so that x² + kx + ___ equals (x + ___)².
Write x² + ___x + ___ as a perfect square.
For y = (x − (___))² + (___), give the vertex.