QuestionFactorise fully $16 - 120k + 225k^{2}$ .

Studdy Solution

STEP 1

Assumptions1. The expression to factorise is $16-120 k+225 k^{}$

STEP 2

The given expression is a quadratic expression of the form $ax^{2} + bx + c$ . In this case, $a =225k^{2}$ , $b = -120k$ , and $c =16$ .

STEP 3

We can factorise a quadratic expression by finding two numbers that add up to $b$ and multiply to $c$ . However, in this case, the quadratic is not in standard form ( $ax^{2} + bx + c$ ). We need to rearrange the terms to put it in standard form.

STEP 4

Rearrange the terms to put the expression in standard form.
$225k^{2} -120k +16$

STEP 5

Now, we need to find two numbers that add up to $b$ and multiply to $c$ . In this case, $b = -120$ and $c =225 \times16$ .

STEP 6

The two numbers that satisfy these conditions are $-30$ and $-30$ because $-30 + -30 = -120$ and $-30 \times -30 =900$ .

STEP 7

Now we can rewrite the middle term of the quadratic expression as the sum of the terms $-30k$ and $-90k$ .
$225k^{2} -30k -90k +16$

STEP 8

Next, we group the terms to prepare for factoring by grouping.
$(225k^{2} -30k) - (90k -16)$

STEP 9

Now, we factor out the greatest common factor from each group.
$15k(15k -2) -4(15k -2)$

STEP 10

Since the terms in parentheses are the same, we can factor them out to get the final factorised form of the expression.
$(15k -2)(15k -4)$ Hence, the factorised form of $16-120 k+225 k^{2}$ is $(15k -2)(15k -4)$ .

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