Question

If ax^3+3x^2+bx-2 has a factor (2x+3) and leaves remainder 7 when divided by (x+2),find the values of a and b and hence factorise the expression​

Answer 1

❏ Question:-

@ If ax^3+3x^2+bx-2 has a factor (2x+3) and leaves remainder 7 when divided by (x+2),find the values of a and b and hence factorise the expression.

❏ Solution:-

➾ Given:-

• f(x)=ax³+3x²+bx-2

• $g_1(x)$=(2x+3)

• $g_2(x)$=(x+2)

✦From the 1'st part of the problem

(ax^3+3x^2+bx-2) has a factor (2x+3)

∴ if f(x) is divided by $g_1(x)$ then it

lefts 0 remainder.

Now, applying Remainder theorem,

➝ $g_1(x)$=2x+3=0

➝ x= $\sf\bf \frac{-3}{2}$

Hence,

➝f( $\sf\bf \frac{-3}{2}$)=a ×$\sf\bf (\frac{-3}{2})^{3}$+3× $\sf\bf (\frac{-3}{2})^{2}$+b ×$\sf\bf \frac{-3}{2}$-2=0

➝a×( $\sf\bf \frac{-27}{8}$)+( $\sf\bf 3\times\frac{9}{4}$)+b×$\sf\bf \frac{-3}{2}$-2=0

➝$\sf\frac{-27a+27\times 2-3b\times 4}{8}=2$

➝$\sf-27a+54-12b=16$

➝$\sf-27a-12b=16-54$

➝$\sf 27a+12b=38$

➝$\sf 12b=38-27a............(i)$

✦From the 2nd part of the problem

∴ if f(x) is divided by $g_2(x)$ then it

lefts remainder=7

Now, applying Remainder theorem,

➝ $g_2(x)$=x+2=0

➝ x=-2

Hence,

$\sf\longrightarrow f(-2)=a(-2)^{3}+3(-2)^{2}+b(-2)-2=7$

$\sf\longrightarrow -8a+12-2b=7+2$

$\sf\longrightarrow -8a-2b=9-12$

[ multiplying by 6 on both sides ]

$\sf\longrightarrow 48a+12b=18$

$\sf\longrightarrow 12b=18-48a..........(ii)$

Now comparing the equations (i) and (ii),

$\sf\longrightarrow 18-48a=38-27a$

$\sf\longrightarrow -48a+27a=38-18$

$\sf\longrightarrow -21a=20$

$\sf\longrightarrow\boxed{\red{ a=\frac{-20}{21}}}$

putting the values of a=$\sf\frac{-20}{21}$ in equation(ii), we get;

$\sf\longrightarrow 12b=18-48(\frac{-20}{21})$

$\sf\longrightarrow 12b=\frac{21\times18+20\times 48}{21}$

$\sf\longrightarrow b=\frac{378+960}{21\times12}$

$\sf\longrightarrow b=\frac{1338}{252}$

$\sf\longrightarrow \boxed{\red{b=\frac{223}{42}}}$

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