What will be the quotient and reminder, if we divide x4-8×2+16 and x+2  using synthetic division?

# What will be the quotient and reminder, if we divide ${x}^{4}-{8x}^{2}+16$ and using synthetic division?

1. A
$\left({{x}^{3}-4x+8\right)}^{}$
2. B
$\left({{x}^{3}-2{x}^{2}-4x+8\right)}^{}$
3. C
$\left({{x}^{3}-2{x}^{2}-x+8\right)}^{}$
4. D
$\left({{x}^{3}-2{x}^{2}+8\right)}^{}$

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### Solution:

According to the question we have to divide ${x}^{4}-{8x}^{2}+16$ and with the help of the synthetic division method. That is  $\frac{{x}^{4}-{8x}^{2}+16}{x+2}$.
And here, ${x}^{4}-{8x}^{2}+16$ is a dividend while is a divisor.
And the dividend is a 4-degree polymer.
We know that the synthetic division method is used when a polynomial is to be divided by a linear expression in which the coefficient should be equal to 1.
So, First, we expand the dividend that is ${x}^{4}-{8x}^{2}+16$ to find its coefficients.
${⇒x}^{4}+{0.x}^{3}-{8x}^{2}+0.x+16$
From here, we can say that ith coefficient of respectively.
Now, we have to find the value of the root of the divisor and for that, we write that the divisor is equal to zero. That is $\left(x+2\right)=0$, and the constant term in the divisor will be the root of $\left(x+2\right)=0$.
So, the constant term in the divisor is:
$\left(x+2\right)=0$
$⇒x=-2$
Now, Bring down the first coefficient of the dividend So, we need to write down the above terms in the following way:

Now, we bring down the first coefficient of the dividend and write the first term directly as following

Now, cross multiply it by -2 and then add it to the above coefficient and proceed with the same process for next all terms

Now, write the above terms as the coefficient of terms starts with the one degree less from the greatest degree which is given in the question as 4.
So, the term would become
${1.x}^{3}+\left(-2\right){.x}^{2}+\left(-4\right){.x}^{1}+8{.x}^{0}+0=\left({x}^{3}-2{x}^{2}-4x+8\right)$
So, the quotient of the division will be and the remainder will be 0, as the term under the constant term after addition is 0
So, we can rewrite it as follows
$\frac{{x}^{4}-{8x}^{2}+16}{x+2}$
$=\left({{x}^{3}-2{x}^{2}-4x+8\right)}^{}+\frac{0}{\left(x+2\right)}$
$=\left({{x}^{3}-2{x}^{2}-4x+8\right)}^{}$
This, the quotient of the division will be $\left({{x}^{3}-2{x}^{2}-4x+8\right)}^{}$and the remainder will be 0.

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