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Factorise fully $ac - ad - bd + bc.$ - Junior Cycle Mathematics - Question b

Question b

Factorise fully $ac - ad - bd + bc.$

Worked Solution & Example Answer:Factorise fully $ac - ad - bd + bc.$ - Junior Cycle Mathematics - Question b

Step 1

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Answer

To factor the expression $ac - ad - bd + bc$ , we can rearrange the terms for clarity. Observe that we can group them as follows:

ac + bc - ad - bd

Next, we can factor by grouping:

(a + b)c - (a + b)d

Now we notice that $(a + b)$ can be factored out:

=(a + b)(c - d)

Step 2

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Answer

Alternatively, we can start from the original expression $ac - ad - bd + bc$ and regroup it as:

ac - bd - ad + bc

Notice that we can rearrange:

(ac + bc) - (ad + bd)

Now, we can factor out common terms from each pair:

(c - d)(a + b)

Thus we reach the same conclusion:

(c - d)(a + b)

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