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.

$a+b=4 \\ ab=1\times 4=4$

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.

$a+b=-8 \\ ab=1\times 16=16$

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.

$x=\frac{-2±\sqrt{2^{2}-4\times 1\times 1}}{2}$

Do the calculations.

$x=\frac{-2±0}{2}$

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.