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bogdanovich [222]
1 year ago
11

" alt="4x {}^{2} + 20x + 25" align="absmiddle" class="latex-formula">
using appropriate identity factories the following​
Mathematics
1 answer:
BigorU [14]1 year ago
3 0

\sf{\qquad\qquad\huge\underline{{\sf Answer}}}

Let's Solve ~

\qquad \sf  \dashrightarrow \: 4 {x}^{2}  + 20x + 25

\qquad \sf  \dashrightarrow \: 4 {x}^{2}  + 10x + 10x + 25

\qquad \sf  \dashrightarrow \: 2x(2x + 5) + 5(2x + 5)

\qquad \sf  \dashrightarrow \: (2x + 5) (2x + 5)

\qquad \sf  \dashrightarrow \: (2x + 5) {}^{2}

or

\sf{\qquad \sf  \dashrightarrow \: 4x² + 20x + 25 }

\sf{\qquad \sf  \dashrightarrow \: (2x)² + (2 \sdot 2x \sdot 5) + (5)²}

[ it's similar to expression - a² + 2ab + b² that is equal to (a + b)² ]

so, let's use this identity here to factorise :

\sf{\qquad \sf  \dashrightarrow \: (2x + 5)²}

I hope it was helpful ~

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we have

<u>The system of equations</u>

y=-3x+6 -------> equation 1

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What are the solutions to the equation?<br> 7x^3=28x
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7x³ = 28x is our equation. We want its solutions.

When you have x and different powers, set the whole thing equal to zero.

7x³ = 28x

7x³ - 28x = 0

Now notice there's a common x in both terms. Let's factor it out.

x (7x² - 28) = 0

As 7 is a factor of 7 and 28, it too can be factored out.

x (7) (x² - 4) = 0

We can further factor x² - 4. We want a pair of numbers that multiply to 4 and whose sum is zero. The pairs are 1 and 4, 2 and 2. If we add 2 and -2 we get zero.

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Now we use the Zero Product Property - if some product multiplies to zero, so do its pieces.

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x - 2 = 0   ----> so x = 2   after adding 2 to both sides

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Thus the solutions are x = 0, x = 2, x = -2.

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