Question

Can you please help me give the Complete solution

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Answer 1

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

• Factorise the polynomials.

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

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1. x² + 4x + 4

`{x}^(2) + 4x + 4`

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

`a+b=4 \\ ab=1* 4=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.

`1,4 \\ 2,2`

Calculate the sum for each pair.

`1+4=5 \\ 2+2=4`

The solution is the pair that gives sum 4.

`a=2 \\ b=2`

Rewrite x² + 4x + 4 as (x² + 2x) + (2x + 4)

`\left(x^(2)+2x\right)+\left(2x+4\right)`

Take out the common factors.

`x\left(x+2\right)+2\left(x+2\right)`

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

`\left(x+2\right)\left(x+2\right)`

Rewrite as a binomial square.

`b. \: \: \boxed{ \boxed{{(x + 2)}^(2) }}`

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2. x² - 8x + 16

`x ^ { 2 } - 8 x + 16`

Factor the expression by grouping. First, the expression needs to be rewritten as x²+ax+bx+16. To find a and b, set up a system to be solved.

`a+b=-8 \\ ab=1* 16=16`

As ab is positive, a and b have the same sign. As a+b is negative, a and b are both negative. List all such integer pairs that give product 16.

`-1,-16 \\ -2,-8 \\ -4,-4`

Calculate the sum for each pair.

`-1-16=-17 \\ -2-8=-10 \\ -4-4=-8`

The solution is the pair that gives sum -8.

`a=-4 \\ b=-4`

Rewrite x²-8x+16 as
`\left(x^(2)-4x\right)+\left(-4x+16\right)`.

`\left(x^(2)-4x\right)+\left(-4x+16\right)`

Take out the common factors.

`x\left(x-4\right)-4\left(x-4\right)`

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

`\left(x-4\right)\left(x-4\right)`

Rewrite as a binomial square.

`d. \: \: \boxed{\boxed{\left(x-4\right)^(2) }}`

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3. 4x² + 12xy + 9y²

`4 x ^ { 2 } + 12 x y + 9 y ^ { 2 }`

Use the perfect square formula,
`a^(2)+2ab+b^(2)=\left(a+b\right)^(2)`, where a=2x and b=3y.

`e. \: \: \boxed{ \boxed{\left(2x+3y\right)^(2) }}`

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4. x⁴ - 2x² + 1

`x ^ { 4 } - 2 x ^ { 2 } + 1`

To factor the expression, solve the equation where it equals to 0.

`x^(4)-2x^(2)+1=0`

By Rational Root Theorem, all rational roots of a polynomial are in the form p/q, where p divides the constant term 1 and q divides the leading coefficient 1. List all candidates p/q.

`± \: 1`

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

`x=1`

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x⁴-2x²+1 by x-1 to get x³+x²-x-1. To factor the result, solve the equation where it equals to 0.

`x^(3)+x^(2)-x-1=0`

By Rational Root Theorem, all rational roots of a polynomial are in the form p/q, where p divides the constant term -1 and q divides the leading coefficient 1. List all candidates p/q.

`± \: \: 1`

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

`x=1`

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x³+x²-x-1 by x-1 to get x²+2x+1. To factor the result, solve the equation where it equals to 0.

`x^(2)+2x+1=0`

All equations of the form ax²+bx+c=0 can be solved using the quadratic formula:
`\frac{-b±\sqrt{b^(2)-4ac}}{2a}`. Substitute 1 for a, 2 for b and 1 for c in the quadratic formula.

`x=\frac{-2±\sqrt{2^(2)-4* 1* 1}}{2} \\`

Do the calculations.

`x=(-2±0)/(2) \\`

Solutions are the same.

`x=-1`

Rewrite the factored expression using the obtained roots.

`\left(x-1\right)^(2)\left(x+1\right)^(2) \\ = a. \: \: \boxed{ \boxed{\left(x^(2)-1\right)^(2)}}`

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• Refer to the attached picture too.