Edumot Online Maths Academy

05/06/2026

QUESTION: Simplify fully (x² + 4)² − (x² − 2)²

SOLUTION: To simply fully, we must factorise the expression completely.

To factorise completely means that we factorise the expression till we come up with an expression that can't be further factotised.

Now let's go!

(x² + 4)² − (x² − 2)²

Note: The expression as a whole comes as a difference of two squares, such as a² – b².

Understand that

a² – b² = (a + b)(a – b)

We use this to factorise the given expression.

(x² + 4)² − (x² − 2)²

= [x² + 4 + (x² − 2)][x² + 4 − (x² − 2)]

Clear the internal brackets

= [x² + 4 + x² − 2][x² + 4 − x² + 2]

Combine like terms

=[2x² + 2][6]

= 6[2x² + 2]

The terms in the bracket has 2 common to each of them. So we factorise that.

= 6[2(x² + 1)]

Multiply 6 by the 2 to get 12

= 12(x² + 1)

This final expression can no longer be factotised. Therefore, we have simplified the given expression fully.

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28/04/2026

QUESTION: Factorise 2y² + xy – 3x²

SOLUTION: To factorise this expression, we modify the middle term without really changing the value of the given expression.

Therefore, we have

2y² + xy – 3x²
= 2y² – 2xy + 3xy – 3x²
= 2y(y – x) + 3x(y – x)
= (y – x)(2y + 3x)

That's it!

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22/04/2026

EDUMOT Online Mathematics Academy

QUESTION: Find y in terms of x in the equation 6log y – 2log x + 3 = 2log y + log x.

SOLUTION: To find y in terms of x is to solve for y or make y the subject of the given equation.

6log y – 2log x + 3 = 2log y + log x
Combine like terms
6log y – 2log y + 3 = log x + 2log x
4log y + 3 = 3log x
4log y + 3log x = 3
log y⁴ + log x³ = 3
log x³y⁴ = 3
Change to index form, we have
x³y⁴ = 10³
x³y⁴ = 10³
Divide both sides by x³
y⁴ = 10³/x³
y⁴ = (10/x)³
Now we have
y = (10/x)³
y = ⁴√(10/x)³

That's it!!!

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09/04/2026

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