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Mila [183]
3 years ago
9

Please help me solve this, will mark as brainliest!!!​

Mathematics
1 answer:
shusha [124]3 years ago
5 0

Answer:

-2 (x-1)^2 + 5

5 units up

1 unit right

Reflection : yes

Stretch

Shrink

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Eight and five hundred seventeen thousandths.
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Help me with solution please =(​
Slav-nsk [51]

\huge \boxed{\mathfrak{Question} \downarrow}

Factorise the polynomials.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

<h3><u>1. b² + 8b + 7</u></h3>

\sf \: b ^ { 2 } + 8 b + 7

Factor the expression by grouping. First, the expression needs to be rewritten as b²+pb+qb+7. To find p and q, set up a system to be solved.

\sf \: p+q=8 \\  \sf pq=1\times 7=7

As pq is positive, p and q have the same sign. As p+q is positive, p and q are both positive. The only such pair is the system solution.

\sf \: p=1  \\  \sf \: q=7

Rewrite \sf\:b^{2}+8b+7 \: as \: \left(b^{2}+b\right)+\left(7b+7\right).

\sf \: \left(b^{2}+b\right)+\left(7b+7\right)

Take out the common factors.

\sf \: b\left(b+1\right)+7\left(b+1\right)

Factor out common term b+1 by using distributive property.

\boxed{\boxed{ \bf \: \left(b+1\right)\left(b+7\right) }}

__________________

<h3><u>2. 4x² + 4x + 1</u></h3>

\sf \: 4 x ^ { 2 } + 4 x + 1

Factor the expression by grouping. First, the expression needs to be rewritten as 4x²+ax+bx+1. To find a and b, set up a system to be solved.

\sf \: a+b=4 \\  \sf ab=4\times 1=4

As ab is positive, a and b have the same sign. As a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

\sf \: 1,4 \\  \sf 2,2

Calculate the sum for each pair.

\sf \: 1+4=5  \\  \sf \: 2+2=4

The solution is the pair that gives sum 4.

\sf \: a=2 \\  \sf b=2

Rewrite \sf4x^{2}+4x+1 as \left(4x^{2}+2x\right)+\left(2x+1\right).

\sf \: \left(4x^{2}+2x\right)+\left(2x+1\right)

Factor out 2x in 4x² + 2x.

\sf \: 2x\left(2x+1\right)+2x+1

Factor out common term 2x+1 by using distributive property.

\sf\left(2x+1\right)\left(2x+1\right)

Rewrite as a binomial square.

\boxed{ \boxed{\bf\left(2x+1\right)^{2}} }

__________________

<h3><u>3. 5n² + 10n + 20</u></h3>

\sf \: 5 n ^ { 2 } + 10 n + 20

Factor out 5. Polynomial n² + 2n+4 is not factored as it does not have any rational roots.

\boxed{\boxed{\bf5\left(n^{2}+2n+4\right)} }

<u>___________</u>_______

<h3><u>4. m³ - 729</u></h3>

\sf \: m ^ { 3 } - 729

Rewrite m³-729 as m³ - 9³. The difference of cubes can be factored using the rule: a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right). Polynomial m²+9m+81 is not factored as it does not have any rational roots.

\boxed{ \boxed{ \bf \: \left(m-9\right)\left(m^{2}+9m+81\right) }}

__________________

<h3><u>5. x² - 81</u></h3>

\sf \: x ^ { 2 } - 81

Rewrite x²-81 as x² - 9². The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

\boxed{ \boxed{ \bf\left(x-9\right)\left(x+9\right) }}

__________________

<h3><u>6. 15x² - 17x - 4</u></h3>

\sf \: 15 x ^ { 2 } - 17 x - 4

Factor the expression by grouping. First, the expression needs to be rewritten as 15x²+ax+bx-4. To find a and b, set up a system to be solved.

\sf \: a+b=-17 \\   \sf \: ab=15\left(-4\right)=-60

As ab is negative, a and b have the opposite signs. As a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

\sf \: 1,-60  \\ \sf 2,-30  \\  \sf3,-20  \\  \sf4,-15 \\   \sf5,-12  \\ \sf 6,-10

Calculate the sum for each pair.

\sf \: 1-60=-59  \\ \sf 2-30=-28 \\   \sf3-20=-17 \\   \sf4-15=-11  \\  \sf5-12=-7  \\  \sf6-10=-4

The solution is the pair that gives sum -17.

\sf \: a=-20  \\  \sf \: b=3

Rewrite \sf15x^{2}-17x-4 as \sf \left(15x^{2}-20x\right)+\left(3x-4\right).

\sf\left(15x^{2}-20x\right)+\left(3x-4\right)

Factor out 5x in 15x²-20x.

\sf \: 5x\left(3x-4\right)+3x-4

Factor out common term 3x-4 by using distributive property.

\boxed{\boxed{ \bf\left(3x-4\right)\left(5x+1\right) }}

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3 years ago
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