QuestionFactor completely: $25 a^{2} - (a - 1)^{2}$ .

Studdy Solution

STEP 1

Assumptions1. We are given the expression $25a^{} - (a-1)^{}$ . . We are asked to factorise it completely.

STEP 2

The given expression can be rewritten as a difference of two squares. The difference of two squares is a term in algebra where it is the subtraction of the squares of two terms. It can be factorised as $(a+b)(a-b)$ .
$25a^{2} - (a-1)^{2}$

STEP 3

We can rewrite $25a^{2}$ as $(5a)^{2}$ and $(a-1)^{2}$ as it is.
$(5a)^{2} - (a-1)^{2}$

STEP 4

Now, we can apply the formula for the difference of two squares, which is $(a+b)(a-b)$ .
$(a + (a-1))(a - (a-1))$

STEP 5

implify the expressions within the brackets.
$(a -1)(4a +1)$ So, the factorised form of the given expression $25a^{2} - (a-1)^{2}$ is $(a -1)(4a +1)$ .

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QuestionFactor completely: $25 a^{2} - (a - 1)^{2}$ .

Studdy Solution

STEP 1

Assumptions1. We are given the expression $25a^{} - (a-1)^{}$ . . We are asked to factorise it completely.

STEP 2

The given expression can be rewritten as a difference of two squares. The difference of two squares is a term in algebra where it is the subtraction of the squares of two terms. It can be factorised as $(a+b)(a-b)$ .
$25a^{2} - (a-1)^{2}$

STEP 3

We can rewrite $25a^{2}$ as $(5a)^{2}$ and $(a-1)^{2}$ as it is.
$(5a)^{2} - (a-1)^{2}$

STEP 4

Now, we can apply the formula for the difference of two squares, which is $(a+b)(a-b)$ .
$(a + (a-1))(a - (a-1))$

STEP 5

implify the expressions within the brackets.
$(a -1)(4a +1)$ So, the factorised form of the given expression $25a^{2} - (a-1)^{2}$ is $(a -1)(4a +1)$ .

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QuestionFactor completely: 25a2−(a−1)225 a^{2} - (a - 1)^{2}25a2−(a−1)2.

Studdy Solution

STEP 1

Assumptions1. We are given the expression 25a−(a−1)25a^{} - (a-1)^{}25a−(a−1). . We are asked to factorise it completely.

STEP 2

The given expression can be rewritten as a difference of two squares. The difference of two squares is a term in algebra where it is the subtraction of the squares of two terms. It can be factorised as (a+b)(a−b)(a+b)(a-b)(a+b)(a−b).25a2−(a−1)225a^{2} - (a-1)^{2} 25a2−(a−1)2

STEP 3

We can rewrite 25a225a^{2}25a2 as (5a)2(5a)^{2}(5a)2 and (a−1)2(a-1)^{2}(a−1)2 as it is.(5a)2−(a−1)2(5a)^{2} - (a-1)^{2} (5a)2−(a−1)2

STEP 4

Now, we can apply the formula for the difference of two squares, which is (a+b)(a−b)(a+b)(a-b)(a+b)(a−b).(a+(a−1))(a−(a−1))(a + (a-1))(a - (a-1)) (a+(a−1))(a−(a−1))

STEP 5

implify the expressions within the brackets.(a−1)(4a+1)(a -1)(4a +1) (a−1)(4a+1)So, the factorised form of the given expression 25a2−(a−1)225a^{2} - (a-1)^{2}25a2−(a−1)2 is (a−1)(4a+1)(a -1)(4a +1)(a−1)(4a+1).

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Studdy solves anything!

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QuestionFactor completely: $25 a^{2} - (a - 1)^{2}$ .

Assumptions1. We are given the expression $25a^{} - (a-1)^{}$ . . We are asked to factorise it completely.

The given expression can be rewritten as a difference of two squares. The difference of two squares is a term in algebra where it is the subtraction of the squares of two terms. It can be factorised as $(a+b)(a-b)$ .
$25a^{2} - (a-1)^{2}$

We can rewrite $25a^{2}$ as $(5a)^{2}$ and $(a-1)^{2}$ as it is.
$(5a)^{2} - (a-1)^{2}$

Now, we can apply the formula for the difference of two squares, which is $(a+b)(a-b)$ .
$(a + (a-1))(a - (a-1))$

implify the expressions within the brackets.
$(a -1)(4a +1)$ So, the factorised form of the given expression $25a^{2} - (a-1)^{2}$ is $(a -1)(4a +1)$ .