Given below is collection of questions related to factorization of polynomial.

All the questions are provided with detailed solution.

The questions are to the standard of grade 09 math.

**Question 01**

Find the value of given polynomial for f (2) and f (-3).

f(x)= \mathtt{2x^{3} -13x^{2} +17x+12}

**Solution**

Let’s first calculate the value of f(2).

Putting the value of 2 in given function.

\mathtt{\Longrightarrow \ 2( 2)^{3} -13( 2)^{2} +17( 2) +12}\\\ \\ \mathtt{\Longrightarrow \ 2( 8) -13( 4) +34+12}\\\ \\ \mathtt{\Longrightarrow \ 16-52+34+12}\\\ \\ \mathtt{\Longrightarrow \ 10}

Hence, value of f(2) is equal to 10.

Now let’s calculate the value of f(-3)

Putting the value of -3 in the given function.

\mathtt{\Longrightarrow \ 2( -3)^{3} -13( -3)^{2} +17( -3) +12}\\\ \\ \mathtt{\Longrightarrow \ 2( -27) -13( 9) -51+12}\\\ \\ \mathtt{\Longrightarrow \ -54-117-51\ +\ 12}\\\ \\ \mathtt{\Longrightarrow \ -210}

Hence, value of f(-3) is equal to -210.

**Question 02**

It’s given that x =2 is the root of following polynomial f(x).

f(x)= \mathtt{2x^{2} -3x+7a}

Find the value of “a” in given function.

**Solution**

Since x = 2 is the root of given function.

The value of function f(x) will be zero when we put 2 in the given expression.

f(2) = 0

\mathtt{2( 2)^{2} -3( 2) +7a=0}\\\ \\ \mathtt{2( 4) -6+7a=\ 0}\\\ \\ \mathtt{8-6+7a\ =\ 0}\\\ \\ \mathtt{7a\ =\ -2}\\\ \\ \mathtt{a=-2/7}

Hence, value of a is -2/7

**Question 03**

if x = -1/2 is the root of below polynomial f(x).

f(x) = \mathtt{8x^{3} -ax^{2} -x+2}

Find the value of variable “a”.

**Solution**

Since x = -1/2 is the root of given expression.

Putting the value of x =-1/2 will result in 0.

\mathtt{8\left( -\frac{1}{2}\right)^{3} -a\left(\frac{-1}{2}\right)^{2} -\left(\frac{-1}{2}\right) +2=0}\\\ \\ \mathtt{8\left(\frac{-1}{8}\right) -a\left(\frac{1}{4}\right) +\frac{1}{2} +2=0}\\\ \\ \mathtt{-1-\frac{a}{4} +\frac{5}{2} =0}\\\ \\ \mathtt{\frac{-4-a+10}{4} =0}\\\ \\ \mathtt{-4-a+10=0}\\\ \\ \mathtt{a=6}

The value of a is 6.

**Question 04**

x= 0 and x = -1 are the root of following polynomial f(x).

f(x) = \mathtt{2x^{3} -3x^{2} +ax+b}

Find the value of a and b.

**Solution**

Since x = 0 is the root, the value of function is 0.

So f(0) = 0

\mathtt{2( 0) -3( 0) +a( 0) +b=0}\\\ \\ \mathtt{b=0}

Hence, value of b = 0.

Also x= -1 is also the root.

f(-1) = 0

\mathtt{2( -1)^{3} -3( -1)^{2} +a( -1) +0=0}\\\ \\ \mathtt{-2-3-a=0}\\\ \\ \mathtt{a\ =-5}

Hence, the value of a = -5.

**Question 05**

Check if x = 1 is the root of below function f(x).

f (x) = \mathtt{\ 9x^{3} -5x^{2} -7x+3}

**Solution**

Put x = 1 in the given function.

\mathtt{\Longrightarrow \ 9( 1)^{3} -5( 1)^{2} -7( 1) +3}\\\ \\ \mathtt{\Longrightarrow \ 9-5-7+3}\\\ \\ \mathtt{\Longrightarrow \ 12-12}\\\ \\ \mathtt{\Longrightarrow \ 0}

Since value of function gets 0, number 1 is the root of the given function.

**Next chapter **: **Problems on factorization using factor theorem**