Mathematics, 17.09.2021 09:20, aliozochika

Help me with solution please =(

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answered: benwill0702

Step-by-step explanation:

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answered: lilay30

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

Factorise the polynomials.

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

1. b² + 8b + 7

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

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.

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

__________________

2. 4x² + 4x + 1

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

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.

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

__________________

3. 5n² + 10n + 20

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

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

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

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4. m³ - 729

$\sf \: m ^ { 3 } - 729$

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.

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

__________________

5. x² - 81

$\sf \: x ^ { 2 } - 81$

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

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

__________________

6. 15x² - 17x - 4

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

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.

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

answered: Guest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$