Name: Introduction to Algebra
Uploaded: 2019-07-19
Duration: 11 min 18 s
Description: In this video, we have discussed basic concepts of Algebra. Having a clear idea of the concept of this chapter will helps you to solve problems which are related to this topic.

Question

$(x - 2)$ is a factor of $(2 x^{3} + 7 x^{2} - 43 x + 42)$ .

Hence, factorise

2 x^{3} + 7 x^{2} - 43 x + 42

completely.

Accepted Answer

If $(x - 2)$ is a factor of $2 x^{3} + 7 x^{2} - 43 x + 42$ , then $2 x^{3} + 7 x^{2} - 43 x + 42$ divide by $(x - 2)$ gives remainder $0$ .

Now using the algebraic division method,

$(2 x^{3} + 7 x^{2} - 43 x + 42) \div (x - 2)$

Divide the first term of the polynomial by $x$ :

$2 x^{3} \div x = 2 x^{2}$ .

Multiply $(x - 2)$ by $2 x^{2}$ : $2 x^{2} (x - 2) = 2 x^{3} - 4 x^{2}$

Subtract: $(2 x^{3} + 7 x^{2}) - (2 x^{3} - 4 x^{2}) = 11 x^{2}$

Bring down the $- 43 x$ from the next column.

$\begin{matrix} 2 x^{2} \\ x - 2 2 x^{3} + 7 x^{2} - 43 x + 42 \\ \frac{2 x^{3} - 4 x^{2}}{11 x^{2} - 43 x} \end{matrix}$

Divide $11 x^{2}$ by $x$ : $11 x^{2} \div x = 11 x$ .

Multiply $(x - 2)$ by $11 x$ : $(11 x) (x - 2) = 11 x^{2} - 22 x$

Subtract: $(11 x^{2} - 43 x) - (11 x^{2} - 22 x) = - 21 x$

Bring down the $42$ from the next column.

$\begin{matrix} 2 x^{2} + 11 x \\ x - 2 2 x^{3} + 7 x^{2} - 43 x + 42 \\ \frac{2 x^{3} - 4 x^{2}}{11 x^{2} - 43 x} \\ \frac{11 x^{2} - 22 x}{- 21 x + 42} \end{matrix}$

Divide $- 21 x$ by $x$ : $- 21 x \div x = - 21$ .

Multiply $(x - 2)$ by $- 21$ : $(- 21) (x - 2) = - 21 x + 42$

Subtract: $(- 21 x + 42) - (- 21 x + 42) = 0$

$\begin{matrix} 2 x^{2} + 11 x - 21 \\ x - 2 2 x^{3} + 7 x^{2} - 43 x + 42 \\ \frac{2 x^{3} - 4 x^{2}}{11 x^{2} - 43 x} \\ \frac{11 x^{2} - 22 x}{- 21 x + 42} \\ \frac{- 21 x + 42}{0} \end{matrix}$

So, $(2 x^{3} + 7 x^{2} - 43 x + 42)$ fully divisible by $(x - 2)$ and gives remainder $0$ .

We know, dividend $=$ divisor $\times$ quotient $+$ remainder

Here, $(2 x^{3} + 7 x^{2} - 43 x + 42) = (x - 2) \times (2 x^{2} + 11 x - 21) + 0$

$\Rightarrow (2 x^{3} + 7 x^{2} - 43 x + 42) = (x - 2) (2 x^{2} + 11 x - 21) \Rightarrow (2 x^{3} + 7 x^{2} - 43 x + 42) = (x - 2) (2 x^{2} + 14 x - 3 x - 21) \Rightarrow (2 x^{3} + 7 x^{2} - 43 x + 42) = (x - 2) (2 x (x + 7) - 3 (x + 7)) \Rightarrow (2 x^{3} + 7 x^{2} - 43 x + 42) = (x - 2) (2 x - 3) (x + 7)$

$(x - 2)$ is a factor of $(2 x^{3} + 7 x^{2} - 43 x + 42)$ .
Hence, factorise $2 x^{3} + 7 x^{2} - 43 x + 42$ completely.

Important Questions on Algebra

Learn and Prepare for any exam you want !

x-2 is a factor of 2x3+7x2-43x+42. Hence, factorise 2x3+7x2-43x+42 completely.

Important Questions on Algebra

Learn and Prepare for any exam you want !

$(x - 2)$ is a factor of $(2 x^{3} + 7 x^{2} - 43 x + 42)$ .
Hence, factorise $2 x^{3} + 7 x^{2} - 43 x + 42$ completely.