Question

Given polynomial $f(x) = {x^3} - 10{x^2} + 19x + 30$ and a factor $x - 6$, how do you find all other factors?

Answer 1

Hint: In this question, we need to find all the factors of a given polynomial when one of the factors is mentioned. Note that we are given a cubic polynomial and a linear factor. So to obtain the other factors, we multiply the linear factor we have with the quadratic polynomial. Then we obtain a cubic polynomial and we equate it with the original polynomial and by comparing coefficients we obtain the required factors of the given polynomial.

Complete step-by-step answer:
Given an equation of the form $f(x) = {x^3} - 10{x^2} + 19x + 30$ …… (1)
And one of the factors $x - 6$.
We are asked to find all other factors of the polynomial given in the equation (1).
Note that the degree of the polynomial given is 3. So it is a cubic polynomial.
And the degree of the factor given is 1, so it is linear.
In order to obtain the other factors, we multiply the linear factor we have, with the quadratic polynomial. So that the resultant polynomial will be of degree 3 and it is cubic polynomial.
Then we equate it with the original cubic polynomial.
Consider the quadratic polynomial of the form $a{x^2} + bx + c$ …… (2)
To find the factors, we need to find the values of a, b and c.
Now we multiply it to the factor $x - 6$, we get,
$ \Rightarrow (x - 6)(a{x^2} + bx + c)$ …(3)
Multiplying each terms, we get,
$ \Rightarrow x(a{x^2}) + bx(x) + xc - 6(a{x^2}) - 6b(x) - 6c$
$ \Rightarrow a{x^3} + b{x^2} + xc - 6a{x^2} - 6bx - 6c$
Combining the like terms together, we get,
$ \Rightarrow a{x^3} + (b - 6a){x^2} + (c - 6b)x - 6c$ ……(4)
Now we equate it to the original polynomial and simplify to obtain the solution.
So equating equations (1) and (4), we get,
${x^3} - 10{x^2} + 19x + 30 = a{x^3} + (b - 6a){x^2} + (c - 6b)x - 6c$
Now we compare the coefficients and solve it.
Comparing the coefficient of ${x^3}$ on both sides of the equation, we get,
$ \Rightarrow a = 1$
Comparing coefficient of ${x^2}$ on both sides of the equation, we have,
$ \Rightarrow - 10 = b - 6a$
We have $a = 1$. So we get,
$ \Rightarrow - 10 = b - 6$
Taking -6 to the other side we get,
$ \Rightarrow - 10 + 6 = b$
$ \Rightarrow b = - 4$
Now comparing the coefficient of x on both sides of the equation, we have,
$ \Rightarrow 19 = c - 6b$
Transferring $ - 6b$ to L.H.S. we get,
$ \Rightarrow 19 + 6b = c$
We have $b = - 4$, so we get,
$ \Rightarrow 19 + 6( - 4) = c$
$ \Rightarrow 19 - 24 = c$
$ \Rightarrow c = - 5$
Comparing the constant term on both sides of the equation, we get,
$ \Rightarrow 30 = - 6c$
We have the value of $c = - 5$
$ \Rightarrow 30 = ( - 6)( - 5)$
$ \Rightarrow 30 = 30$
Substituting the values of a, b, c in the equation (2), we get,
$ \Rightarrow a{x^2} + bx + c = 1{x^2} + ( - 4)x + ( - 5)$
$ \Rightarrow {x^2} - 4x - 5$
We can write the middle term as, $ - 4x = x - 5x$. So we get,
$ \Rightarrow {x^2} + x - 5x - 5$
Taking out the common terms we get,
$ \Rightarrow x(x + 1) - 5(x + 1)$
$ \Rightarrow (x - 5)(x + 1)$
Hence the equation (3) becomes,
$ \Rightarrow (x - 6)(x - 5)(x + 1)$
Therefore, finally we have, $f(x) = {x^3} - 10{x^2} + 19x + 30 = (x - 6)(x - 5)(x + 1)$.

Note:
If we have given a cubic polynomial along with one of the factors and asked to find out the all other factors, we need to multiply the given factor which will be in linear form with the quadratic polynomial. Then we obtain a cubic polynomial and we equate it with the original polynomial. To find the factors we must compare the coefficients on both sides of the polynomials. Students must be careful while comparing. There may be chances where they go wrong while comparing. Terms with the same power have to compare on both sides and obtain the solution.