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Question: How do you factor \({{x}^{2}}-3x+15\) ?...

How do you factor x2−3x+15{{x}^{2}}-3x+15 ?

Explanation

Solution

In this question, we have to find the factors of the equation. The equation given to us is in the form of a quadratic, therefore when we solve this problem, we will get two values for x, which satisfy the equation. Therefore, we will apply the discriminant method to solve this problem. We compare the general form of quadratic equation and the given equation to get the value of a, b, and c. Then, we will get the value of discriminant D=b2−4acD=\sqrt{{{b}^{2}}-4ac}, and thus find the value of x using the discriminant formula x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} .After necessary calculations, get two equations of x, we solve them separately to get the value of x, which is our required answer.

Complete step-by-step solution:
According to the question, a quadratic equation is given to us and we have to solve the equation to get the factor for the same.
The equation is x2−3x+15{{x}^{2}}-3x+15 ----------------- (1)
As we know, the general quadratic equation is in form of ax2+bx+c=0a{{x}^{2}}+bx+c=0 ---------- (2)
Thus, on comparing equation (1) and (2), we get a=1,a=1, b=−3,b=-3, and c=15c=15 ------- (3)
So, now we will apply the discriminant formula D=b2−4acD=\sqrt{{{b}^{2}}-4ac} by putting the above values in the formula, we get
⇒D=(−3)2−4.(1).(15) ⇒D=9−60 \begin{aligned} & \Rightarrow D=\sqrt{{{(-3)}^{2}}-4.(1).(15)} \\\ & \Rightarrow D=\sqrt{9-60} \\\ \end{aligned}
Thus, on further solving, we get
⇒D=−51\Rightarrow D=\sqrt{-51}
Since D<0 implies it has no real roots which means it has complex roots,therefore we now use the iota to get rid of the negative square root, we get
⇒D=±i51\Rightarrow D=\pm i\sqrt{51} -------------- (4)
Since we see the discriminant is a non-real number, thus now we will find the value of x, using the formula,
⇒x=−b±D2a\Rightarrow x=\dfrac{-b\pm D}{2a}
⇒x=−b±b2−4ac2a\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} --------------- (5)
So, we will put the value of equation (3) and (4) in equation (5), we get
⇒x=−(−3)±i512.(1)\Rightarrow x=\dfrac{-(-3)\pm i\sqrt{51}}{2.(1)}
On further simplification, we get
⇒x=3±i512\Rightarrow x=\dfrac{3\pm i\sqrt{51}}{2}
Therefore, we will split the above equation in terms of (+) and (-), we get
⇒x=3+i512\Rightarrow x=\dfrac{3+i\sqrt{51}}{2} -------- (6) , or
⇒x=3−i512\Rightarrow x=\dfrac{3-i\sqrt{51}}{2} ---------- (7)
Now, we will first solve equation (6), we get
⇒x=32+i512\Rightarrow x=\dfrac{3}{2}+\dfrac{i\sqrt{51}}{2}
Now we will solve equation (7), we get
⇒x=32−i512\Rightarrow x=\dfrac{3}{2}-\dfrac{i\sqrt{51}}{2}
Therefore, for the equation x2−3x+15{{x}^{2}}-3x+15 , we get the value ofx=32+i512,32−i512x=\dfrac{3}{2}+\dfrac{i\sqrt{51}}{2},\dfrac{3}{2}-\dfrac{i\sqrt{51}}{2}.

Note: While solving this problem, do all the steps carefully and avoid errors to get the correct answer. One of the alternative methods for solving this problem is using splitting the middle term method. In this method, we first split the middle term in the factors of the constant value and then take the common value among all and make necessary calculations to get the required value of x.