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Question

Question: How do you factor \(8{{x}^{2}}+65x+8\)?...

How do you factor 8x2+65x+88{{x}^{2}}+65x+8?

Explanation

Solution

We use both grouping method and vanishing method to find the factor of the problem. We take common terms out to form the multiplied forms. In the case of the vanishing method, we use the value of x which gives the polynomial value 0.

Complete step-by-step solution:
We apply the middle-term factoring or grouping to factorise the polynomial.
Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number.
In the case of 8x2+65x+88{{x}^{2}}+65x+8, we break the middle term 65x65x into two parts of 64x64x and xx.
So, 8x2+65x+8=8x2+64x+x+88{{x}^{2}}+65x+8=8{{x}^{2}}+64x+x+8. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases gives 64x264{{x}^{2}}. The grouping will be done for 8x2+64x8{{x}^{2}}+64x and x+8x+8. We try to take the common numbers out.
For 8x2+64x8{{x}^{2}}+64x, we take 8x8x and get 8x(x+8)8x\left( x+8 \right).
For x+8x+8, we take 1 and get (x+8)\left( x+8 \right).
The equation becomes 8x2+65x+8=8x2+64x+x+8=8x(x+8)+(x+8)8{{x}^{2}}+65x+8=8{{x}^{2}}+64x+x+8=8x\left( x+8 \right)+\left( x+8 \right).
Both the terms have (x+8)\left( x+8 \right) in common. We take that term again and get
8x2+65x+8 =8x(x+8)+(x+8) =(x+8)(8x+1) \begin{aligned} & 8{{x}^{2}}+65x+8 \\\ & =8x\left( x+8 \right)+\left( x+8 \right) \\\ & =\left( x+8 \right)\left( 8x+1 \right) \\\ \end{aligned}
Therefore, the factorisation of 8x2+65x+88{{x}^{2}}+65x+8 is (x+8)(8x+1)\left( x+8 \right)\left( 8x+1 \right).

Note: We find the value of x for which the function f(x)=8x2+65x+8=0f\left( x \right)=8{{x}^{2}}+65x+8=0. We can see f(8)=8×(8)2+65×(8)+8=512520+8=0f\left( -8 \right)=8\times {{\left( -8 \right)}^{2}}+65\times \left( -8 \right)+8=512-520+8=0. So, the root of the f(x)=8x2+65x+8f\left( x \right)=8{{x}^{2}}+65x+8 will be the function (x+8)\left( x+8 \right).
This means for x=ax=a, if f(a)=0f\left( a \right)=0 then (xa)\left( x-a \right) is a root of f(x)f\left( x \right).
Now, f(x)=8x2+65x+8=(x+8)(8x+1)f\left( x \right)=8{{x}^{2}}+65x+8=\left( x+8 \right)\left( 8x+1 \right). We can also do this for (8x+1)\left( 8x+1 \right).