Question

Please solve in this question

Using Factor Theorem, Factorize x ^ 3 + 10x ^ 2 - 37x + 26​

Answer 1

$\small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

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Factor the following.

$\rm{x}^{3}+10{x}^{2}-37x+26$

2 First, find all factors of the constant term 26.

1, 2, 13, 26

Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is 0, x-1 is a factor..

$\rm{1}^{3}+10\times {1}^{2}-37\times 1+26 = 0$

• x-1

Polynomial Division: Divide $\rm{x}^{3}+10{x}^{2}-37x+26$

[NOTE]:-

$\fcolorbox{green}{lightyellow}{\boxed{\scriptsize{\mathbb\pink{REFERR \:TO\: THE\:\: ATTACHMENT }}}}$

Rewrite the expression using the above.

$\rm({x}^{2}+11x-26)(x-1)$

Factor x²+11x-26.

1.Ask: Which two numbers add up to 11 and multiply to −26?

−2 and 13

2. Rewrite the expression using the above.

(x-2)(x+13)

$\fcolorbox{magenta}{olive}{ \boxed{\red{ \rm \large{\dashrightarrow} \small(x-2)(x+13)(x-1)}}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: {x}^{3} + {10x}^{2} - 37x + 26$

Let assume that

$\rm \: f(x) = {x}^{3} + {10x}^{2} - 37x + 26$

Let we first find the factor of f(x) by using hit and trial method.

Let assume that x = 1, so we get

$\rm \: f(1) = {(1)}^{3} + {10(1)}^{2} - 37(1) + 26$

$\rm \: = \: 1 + 10 - 37 + 26$

$\rm \: = \: 37 - 37$

$\rm \: = \: 0$

$\rm\implies \:x - 1 \: is \: a \: factor \: of \: f(x)$

So, by using Long Division Method, we have

$\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{\underline{\sf{\:\: {x}^{2} + 11x - 26\:\:}}}\\ {\underline{\sf{x - 1}}}& {\sf{\: {x}^{3} +   {10x}^{2} - 37x + 26\:\:}} \\{\sf{}}& \underline{\sf{ \:- {x}^{3} + {x}^{2} \: \: \: \: \: \: \: \: \: \:    \:  \:  \:  \:  \:  \:  \: \:\:}} \\ {{\sf{}}}& {\sf{\: \: \: \: \: \: \: \: \: 11{x}^{2} - 37x + 26 \: }} \\{\sf{}}& \underline{\sf{ \: \: \: \: \: \: \: \: \: - 11{x}^{2} + 11x  \:  \:  \:  \:  \:  \: \:\:}} \\ {\underline{\sf{}}}& {\sf{\:\:   - 26x + 26  \:\:}} \\{\sf{}}& \underline{\sf{\: \: \: \: \: \: \: 26x - 26\:\:}} \\ {\underline{\sf{}}}& {\sf{ \: \: \: \: \: \: \: \: 0\:\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered}$

We know,

Dividend = Divisor × Quotient + Remainder

So, on substituting the values from above, we have

$\rm \: {x}^{3} + {10x}^{2} - 37x + 26 = (x - 1)( {x}^{2} + 11x - 26)$

$\rm \: = (x - 1)( {x}^{2} + 13x - 2x - 26)$

$\rm \: = (x - 1)[x(x + 13) - 2(x + 13)]$

$\rm \: = (x - 1)[(x + 13)(x - 2)]$

$\rm \: = \: (x - 1)(x - 2)(x + 13)$

Hence,

$\red{\boxed{ \rm{ \:\sf \: {x}^{3} + {10x}^{2} - 37x + 26 = (x - 1)(x - 2)(x + 13)} \: }}$

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Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$