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4 years ago

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What is -26b4 + (-2b4) Simplified

2 answers:

Pepsi [2]4 years ago

5 0

Hello,

As I simplified the expression, The answer is -28b^4

I hope this helps

denis23 [38]4 years ago

4 0

<span><span>−<span>26<span>b4</span></span></span>−<span>2<span>b4</span></span></span><span>=<span><span><span>−<span>26<span>b4</span></span></span>+</span>−<span>2<span>b4</span></span></span></span><span>=<span><span>−<span>26<span>b4</span></span></span>+<span>−<span>2<span>b4</span></span></span></span></span><span>=<span>(<span><span>−<span>26<span>b4</span></span></span>+<span>−<span>2<span>b4</span></span></span></span>)</span></span><span>=<span>−<span>28<span>b4</span></span></span></span>

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Can someone please help me with this question?

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3 years ago

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ExtremeBDS [4]

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Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

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So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

<em><u>Therefore, the two roots are:</u></em>

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

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3 years ago

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7 0

4 years ago