The expression inside the parentheses matches the algebraic identity x² - 2xy + y² = (x - y)², where x = 2b and y = 7c. Checking the middle term, 2 × 2b × 7c = 28bc, which confirms it is a perfect square trinomial.

The expression inside the parentheses matches the algebraic identity x² - 2xy + y² = (x - y)², where x = 2b and y = 7c. Checking the middle term, 2 × 2b × 7c = 28bc, which confirms it is a perfect square trinomial.

We rewrite 25a² as (5a)². The entire expression is now of the form X² - Y², where X = 5a and Y = (2b - 7c).

25 a square minus 4b square + 28 BC - 49 c square and factorise that

Answer: $(5a + 2b - 7c)(5a - 2b + 7c)$

Step-by-step solution

Step 1: Group the terms involving b and c

Observe that the last 3 terms contain variables $b$ and $c$ , along with the cross product term $28bc$ . We can factor out a negative sign from these 3 terms to form a grouped quadratic trinomial in parentheses.

Step 2: Express the grouped trinomial as a perfect square

The expression inside the parentheses matches the algebraic identity $x^2 - 2xy + y^2 = (x - y)^2$ , where $x = 2b$ and $y = 7c$ . Checking the middle term, $2 \times 2b \times 7c = 28bc$ , which confirms it is a perfect square trinomial.

Step 3: Rewrite the expression as a difference of two squares

We rewrite $25a^2$ as $(5a)^2$ . The entire expression is now of the form $X^2 - Y^2$ , where $X = 5a$ and $Y = (2b - 7c)$ .

Step 4: Apply the difference of squares identity

Applying the identity $X^2 - Y^2 = (X + Y)(X - Y)$ , we substitute $X = 5a$ and $Y = 2b - 7c$ . Expanding the brackets carefully gives $(5a + 2b - 7c)$ and $(5a - 2b + 7c)$ .

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