\( \begingroup \)Question 1 a) The expression $\var{pmult*pxcoeff}x+\var{pconstant*pmult}$ is a sum and can be factorised (written as a product) by finding the largest common factor: $\var{pmult*pxcoeff}x+\var{pconstant*pmult} = $ Expected answer: $\large($ interpreted asExpected answer:interpreted as5x+7 $\large)$ We put the common factor out the front of a set of brackets and put the 'left-overs' inside. The (largest) common factor of $\var{pmult*pxcoeff}x+\var{pconstant*pmult}$ is $\var{pmult}$. Once we remove that factor from each term in $\var{pmult*pxcoeff}x+\var{pconstant*pmult}$ we are left with $\var{pxcoeff}x+\var{pconstant}$. That means $\var{pmult*pxcoeff}x+\var{pconstant*pmult}= \var{pmult}(\var{pxcoeff}x+\var{pconstant})$. Submit partYour answer is being marked. Please wait. This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback. Not marked What do you want to do next? ⤺ Go back to the previous part There's nothing more to do from here. Show stepsHide steps(You will lose 1 mark.) Submit partYour answer is being marked. Please wait. This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback. Score: 0/2 Unanswered Or, you could: ⤺ Go back to the previous part There's nothing more to do from here.b) Factorise $\simplify{{bp1}a+{bp2}}$ Expected answer: $\large($ interpreted asExpected answer:interpreted as $\large)$ We put the common factor out the front of a set of brackets and put the 'left-overs' inside. The (largest) common factor of $\simplify{{bp1}a+{bp2}}$ is $\var{cf}$. Once we remove that factor from each term in $\simplify{{bp1}a+{bp2}}$ we are left with $\var{bx}a+\var{bc}$. That means $\simplify{{bp1}a+{bp2}}$ is $\var{cf} = \var{cf}(\var{bx}a+\var{bc})$. Submit partYour answer is being marked. Please wait. This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback. Not marked What do you want to do next? ⤺ Go back to the previous part There's nothing more to do from here. Show stepsHide steps(You will lose 1 mark.) Submit partYour answer is being marked. Please wait. This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback. Score: 0/2 Unanswered Or, you could: ⤺ Go back to the previous part There's nothing more to do from here.c) Factorise $\simplify{{ct1}x+{ct2}y+{ct3}}$ Expected answer: $\large($ interpreted asExpected answer:interpreted as $\large)$ We put the common factor out the front of a set of brackets and put the 'left-overs' inside. The (largest) common factor of $\simplify{{ct1}x+{ct2}y+{ct3}}$ is $\var{cmult}$. Once we remove that factor from each term in $\simplify{{ct1}x+{ct2}y+{ct3}}$ we are left with $\simplify{{cx}x+{cy}y+{cc}}$. That means $\simplify{{ct1}x+{ct2}y+{ct3}} = \simplify{{cmult}({ct1}x+{ct2}y+{ct3})}$. Submit partYour answer is being marked. Please wait. This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback. Not marked What do you want to do next? ⤺ Go back to the previous part There's nothing more to do from here. Show stepsHide steps(You will lose 1 mark.) Submit partYour answer is being marked. Please wait. This feedback is based on your last submitted answer. Submit your changed answer to get updated feedback. Score: 0/2 Unanswered Or, you could: ⤺ Go back to the previous part There's nothing more to do from here. Advice \( \endgroup \)